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1 Problems Ted Eisenberg, Section Editor ********************************************************* This section of the Journal offers readers an opportunity to echange interesting mathematical problems and solutions. Please send them to Ted Eisenberg, Department of Mathematics, Ben-Gurion University, Beer-Sheva, Israel or fa to: Questions concerning proposals and/or solutions can be sent to Solutions to previously stated problems can be seen at < Solutions to the problems stated in this issue should be posted before October 5, : Proposed by Kenneth Korbin, New York, NY Given a triangle with sides (,, 4). The sum of these digits is =. Find primitive pythagorean triples in which the sum of the digits is or less. 55: Proposed by Moshe Stupel, Shaanan Academic College of Education and Gordon Academic College of Education, and Avi Sigler, Shaanan Academic College of Education, Haifa, Israel Without the use of a calculator, show that: 8 sin sin 4 sin 6 sin 8 =. 55: Proposed by D.M. Bătinetu-Giurgiu, Bucharest, Romania, Neculai Stanciu, George Emil Palade School Buzău, Romania and Titu Zvonaru, Comănesti, Romania Determine all real numbers a, b,, y that simultaneously satisfy the following relations: () a + by = 5 () a + by = 9 () a + by = 7 (4) a 4 + by 4 =. 55: Proposed by Daniel Sitaru, Theodor Costescu National Economic College, Drobeta Turnu-Severin, Mehedinti, Romania Prove that if a, b, c > and a + b + c = e then e ace e bae e cbe > e e a be b ce c ae. ( Here, e = lim + n n n)

2 55: Proposed by José Luis Díaz-Barrero, Barcelona Tech, Barcelona, Spain Let a, a,..., a n be positive real numbers with n. Prove that where m is a positive integer. (a m a + a m a + + a m n a ) m (a m + am + am n ) m+ n, 554: Proposed by Ovidiu Furdui and Alina Sîntămărian both at the Technical University of Cluj-Napoca, Cluj-Napoca, Romania Let n be an integer. Calculate where denotes the integer part of. n d, Solutions 548: Proposed by Kenneth Korbin, New York, NY A triangle with integer area has integer length sides (,, + ). Find five possible values of with > 4. Solution by Dionne Bailey, Elsie Campbell, Charles Diminnie, and Trey Smith, Angelo State University, San Angelo, TX For our approach, we will need to find positive integer solutions for the equation One way to do so is to first solve the Pell Equation and then set m = X and k = Y. m 8k = 9. () X 8Y = () Following the usual process for solving (), we note that the solution with the smallest X value is X =, Y =. Then, all solutions (X n, Y n ) of () can be found by setting X n + Y n 8 = ( + 8 for all n. Then, as described above, we get solutions for () by setting m n = X n and k n = Y n. The first si solutions for () and () are listed in the following table: ) n n X n Y n m n k n , 7 6 5, 6, 89, 89, , 6 6, 9 58, 8, 79 ()

3 For the problem at hand, the semiperimeter s of our triangle is s = + + ( + ) and Heron s Formula for the area A yields = + A = s (s ) (s ) (s ) = ( + ) ( ) () () = ( + ). For A to be a positive integer, we must find a positive integer k for which or A = ( + ) = 4k By the Quadratic Formula, the positive solution of (4) is For to be a positive integer, we will need or + k =. (4) = (k + ) = + 8k k + 9 = m m 8k = 9 for some odd positive integer m. However, table () gives us si solutions to use. In each case, = + 8k + 9 = m and A = k. The solution m = 9 and k = yields = 4, which is ruled out in the statement of the problem. The other five entries in the table provide five plausible values of for which A is a positive integer. These values are listed in our final table: n m n k n A , , 4 5, 89, 567 5, 44 7, , 8, 79 9, 4 4, 58 Solution by Bruno Salgueiro Fanego, Viveiro, Spain

4 Let p and S be the semi perimeter and the area of such a triangle respectively. Then p = = + 4 and, by Heron s formula S = p(p )(p )(p ( + )) = + 4 must be an integer. It can be easily verified that for each of the five values of {5, 48, 865, 544, 94} one obtains triangles that have areas of 6,, 4, 74, 458, respectively. ( ) ( ) 5 More generally, if =, then the recurrence given by S 6 ( ) ( ) ( ) ( ) ( ) n+ n n+ = + for any integer n, gives a pair S n+ 4 S n S n+ where n+ is the length of a triangle with integer length sides (, n+, n+ + ) and S n+ is the integer area of that triangle. Solution by Julio Cesar Mohnsam and Luiz Lemos Junior, both at IFSUL Campus Pelotas-RS, Brazil Let p be the semi-perimeter p = = + The area by Heron is given by: A = p(p )(p )(p ) = ( + )( )()() Then ( + )( )() must be a square, that is, + 4 = y, follow that: Multiplying () by 8 we have: + y 4 = () Adding 4 on both sides of () we have: Now make X = 4 + and Y = y, we have: y = () (4 + ) 8y 6 = () X 8Y 6 = (4) (4) is a diophantine equation of the form a by + c = in the case of c = we have the particular case of the equation of Pell Dy =. If (a, b) c, the equation has a solution. Let s solve (4) using the method Florentin Smarandache []. We consider the equation and ax by + c = (5) We set the matri A from (6) as follows: [ b α A = a β ] β α where (α, β ) are initial solutions of (6). aα bβ = a (6) 4

5 Now let (X, Y ) are initial solutions of (5), then the general solutions of (5) are given by the following recurrence relation: [ ] [ ] Xn = A n X Y Y n Thus, by solving (4) we have to first assemble matri A from X 8Y =, note that this Pell equation has initial solution (, ), so we have: [ ] 8 A = But (4) has initial solution (6, ). So we have to: [ ] [ ] [ X 8 6 = Y ] = [ 8 6 Like 8 = X = 4 + = 4 but we have to find > 4. Thus we calculate A, such that: [ X Y ] = [ 8 ] [ 6 ] = [ ] [ 6 ] ] = [ 6 As = X = 4 + = 5 and the lengths of the first triangle are (, 5, 6). To find the other values of we will diagonalize the matri A. We know that A n = P D n P. [ ] A = P DP λ = P P λ Eigenvalues λ = + and λ = ( ) ( ) Eigenvectors v = and v = Therefore = {5, 48, 565, 544, 94} [] Smarandache F. Un metodo de resolucion de la ecuacion diofantica. Gazeta Matematica, Serie, Vol., Nr., 988. Madrid. p Solution 4 by Brian D. Beasley, Presbyterian College, Clinton, SC Given such a triangle, its semiperimeter is s = ( )/ = +. Then by Heron s formula, its area is = (( + )( )()() = ( + ), so we seek integers and with > 4 such that = + 4. This equation in turn is equivalent to ( ) ( ) = ( + ), or + =. We let a = ( + )/ and b = / in order to solve the Pellian equation b + = a. This equation has infinitely many integer solutions for a and b, which we may describe with the sequences a =, a =, and a n+ = 6a n+ a n for n ; b =, b =, and b n+ = 6b n+ b n for n. 5 ]

6 Thus there are infinitely many integers which satisfy the requirements of the problem, given by the terms n (with n ) of the sequence =, = 4, and n+ = 6 n+ n + for n. In particular, the net five values of after 4 are 5, 48, 865, 544, and 94. Addenda. (i) We may also describe the above sequences by letting γ = + and δ =. Then a n = (γ n + δ n )/ for each n, which implies that n = (γ n + δ n )/4. (ii) We further note that the ratios a n /b n for n occur as every other term in the sequence of converging to the continued fraction representation of. Comments by Editor : Ioannis D. Sfikas of Athens Greece started his solution off with some nomenclature and bit of history about the problem. A triangle whose sides and area are rational numbers is called a rational triangle. If the rational triangle is right-angled, it is called a right-angled rational triangle or a rational Pythagorean triangle or a numerical right triangle. If the sides of a rational triangle is of integer length, it is called an integer triangle. If further these sides have no common factor greater than unity, the triangle is called a primitive integer triangle. If the integer triangle is right-angled, it is called a Pythagorean triangle. A Heronian triangle (named after Heron of Aleandria) is an integer triangle with the additional property that its area is also an integer. A Heronian triangle is called primitive Heronian triangle if sides have no common factor greater than unity. In the 7th century, the Indian mathematician Brahmagupta studied the special case of triangles with consecutive integer sides. Kenneth Korbin, the proposer of this problem stated that triangles sides with lengths (,, + ) with 4 have an area of + 4, and are associated with the sequences of (5, 48, 865, 544, 94,..., N,...) that satisfies the recursion of N+ = 6 N N +. David Stone and John Hawkins of Southern Georgia University asked in their solution, why the values of 4? Why not >? They then stated that: If =, the triangle (,, ) is degenerate; =, the triangle (,, ) has area 8 and is not Heronian; =, the triangle (,, 4) has area and is not Heronian; and = 4, the right triangle (, 5, 5) is too easy. They then asked: What about those Heronian triangles of the form (,, + ), where is an integer. Applying Heron s Formula they obtained that 6A = and stated that there are no integer solutions to this for because the left-and side is even while the right hand side is odd. They then looked at triangles of the form (,, + ) and stated that the only triangle of this form is degenerate. Moreover, no triangle of the form (,, + d) can eist for d >. They continued on with the following: Thus the problem poser selected the one form that does admit solutions. Still to be 6

7 investigated: finding the Heronian triangles of the form (4,, + ), and those of the form (4,, + ), etc. Their solution ended with the statement: There are no Heronian triangles of the form (n,, + d) for positive integers n and d having opposite parity. Also solved by Hatef I. Arshagi, Guilford Technical Community College, Jamestown, NC; Jeremiah Bartz, University of North Dakota, Grand Forks, ND; Anthony Bevelacqua University of North Dakota, Grand Forks, ND; Ed Gray, Highland Beach, FL; Paul M. Harms, North Newton, KS; Kee-Wai Lau, Hong Kong, China; Carl Libis, Columbia Southern University, Orange Beach, AL; David E. Manes, Oneonta, NY; Ioannis D. Sfikas, National and Kapodistrian University of Athens, Greece; Albert Stadler, Herrliberg, Switzerland; Neculai Stanciu, George Emil Palade School, Buzău and Tito Zvonaru, Comănesti, Romania; David Stone and John Hawkins of Southern Georgia University, Statesboro, GA, and the proposer. 548: Proposed by Daniel Sitaru, Theodor Costescu National Economic College, Drobeta Turnu-Severin, Mehedinti, Romania Prove that if n is a natural number then (tan5 ) n (tan4 ) n + (tan ) n + (tan4 ) n (tan ) n + (tan ) n + (tan ) n (tan ) n + (tan ) n. Solutions and by Henry Ricardo, Westchester Area Math Circle, NY Solution. Since, for a fied natural number n, (tan ) n is an increasing positive function for [, 9 ), we have (tan 5 ) n (tan 4 ) n + (tan ) n (tan 4 ) n (tan ) n + (tan ) n (tan ) n (tan ) n + (tan ) n (tan 5 ) n (tan 5 ) n + (tan 5 ) n =, (tan 4 ) n (tan 4 ) n + (tan 4 ) n =, (tan ) n (tan ) n + (tan ) n =, so that adding these inequalities gives us the desired result. Equality holds if and only if n = (assuming that is considered a natural number). Solution. Since, for a fied natural number n, (tan ) n is an increasing positive function for [, 9 ), we have (tan ) n (tan ) n + (tan ) n and (tan 4 ) n (tan ) n + (tan ) n 7 (tan ) n (tan 4 ) n + (tan 5 ) n, (tan 4 ) n (tan ) n + (tan 5 ) n,

8 so that (tan ) n (tan ) n + (tan ) n cyclic (tan 5 ) n (tan 4 ) n + (tan ) n + (tan 4 ) n (tan ) n + (tan 5 ) n + (tan ) n (tan 4 ) n + (tan 5 ) n. Setting a = (tan ) n, b = (tan 4 ) n, and c = (tan 5 ) n, we see that the right-hand side of the last inequality has the form a b + c + b c + a + c a + b, for a, b, c >, which is greater than or equal to / by Nesbitt s inequality. Equality holds if and only if n = (assuming that is considered a natural number). Solution by Ed Gray, Highland Beach, FL First we retrieve the required values:. tan = tan = tan = tan 4 = tan 5 = We rewrite the problem s equation as: tan 4 + tan + tan 5 tan 5 tan + tan + tan 4 tan 4 tan tan + tan tan Substituting the values from steps -5 and performing the indicated divisions we define: f(n) = ( ) n + (.59965) n + ( ) n + (.49946) n + ( ) n + (.648) n. We note that f(n) is an increasing function of n since the denominators clearly decrease as n increases. Finally we note that f() = = >. Then the equality holds for all n since f(n) is an increasing function. Solution 4 by David Stone and John Hawkins, Georgia Southern University, Statesboro, GA Lemma: For fied positive reals a, b, c with a < c, b < c let f() = Then f(), for, with equality holding only for =. Proof: We calculate the derivative: c b for. + a f () = (b + a ) c ln c c (a ln a + b ln b) (b + a ) 8

9 = c (b + a ) ln c (a ln a + b ln b) (b + a ) = c b (ln c ln b) + a (ln c ln a) (b + a ). The ln function is increasing, so ln c > ln b and ln c > ln a; thus we see that the derivative is positive. Hence the function f is increasing, so = f() f() for. Because the derivative is strictly positive, the function f actually grows: so f() > for >. To verify the inequality of the problem, we note that the tangent function is increasing, so in each summand the tangent term in the numerator is larger that each tangent term in the denominator. Hence we can apply the lemma to each of the three summands, forcing the sum. Note that equality holds if and only if n =. Comment: We can apply the lemma to obtain some ugly inequalities which are clearly true: n n + n + 4n n + n + 5n (n + )n n n n n + (n + ) n n, and [(n + )!] n [n!] n + [(n + )!] n. Also solved by Arkady Alt, San Jose, CA (two solutions); Paul M. Harms, North Newton, KS; Kee-Wai Lau, Hong Kong, China; Paolo Perfetti, Department of Mathematics, University of Tor Vergata, Rome, Italy; Angel Plaza, University of Las Palmas de Gran Canaria, Spain; Ioannis D. Sfikas, National and Kapodistrian University of Athens, Greece; Albert Stadler, Herrliberg, Switzerland, and the proposers 548: Proposed by D.M. Bătinetu-Giurgiu, Matei Basarab National College, Bucharest and Neculai Stanciu, George Emil Palade School Buzău, Romania ( If a, b >, and, π ) then show that (i) (ii) (a + b) sin + ab a + b tan 6ab a + b. a tan + b sin > ab. Solution by Paolo Perfetti, Department of Mathematics, Tor Vergata University, Rome, Italy Proof of (i). 9

10 The AHM yields and then Thus we prove This is equivalent to (a + b) sin a + b 4 a + b (a + b) 4ab + ab a + b tan 4ab a + b sin + ab a + b tan 4ab ab 6ab sin + tan a + b a + b a + b f(). = 4 sin + tan 6 f () = 4 cos + cos 6 f () = 4 sin + 4 sin cos = 4 sin ( 4 cos ) > via cos (, ) for < < π/. Since f () = f() = we get f(). Proof of (ii). Let f() = a tan + b sin ab, f() = f () = a cos + b cos ab and this concludes the proof. Solution by Arkady Alt, San Jose, CA (i) First we will prove inequality tan + sin > tan + sin a cos + b cos ab >, (, π/). ab cos cos ab = Let h () := tan + sin, (, π/). Since h () = cos + cos = ( cos + ) ( cos ) cos >, (, π/) then h () > h () =. Hence, (a + b) sin + ab a + b tan > (a + b) sin + ab a + b ( sin a + b 4ab ) + 6ab a + b a + b = sin (a b) + 6ab a + b a + b 6ab a + b. ( sin ) = (ii) Let h () := a tan + b sin ab. Since h () = a cos + b cos ab a cos b cos ab = ab cos > then h () > h () = cos a tan + b sin > ab. Solution by Kee-Wai Lau, Hong Kong, China

11 ( It is well known that for, π ), we have sin 6 4ab + (a b) a + b = ab a + b a + b ( ) sin + tan and tan +. Since 4ab, so the left side of (i) is greater than or equal to a + b ab ( a + b ) ( + 6 )) ( + = 6ab a + b, as required. ( It is also well known that for, π ), cos + 4, so that 4 ) ) sin = cos ( ( = 4 ( ) 4 7. Hence, and (ii) holds. a tan + b sin (a tan )(b sin ) = sin ab cos ab Also solved by Ioannis D. Sfikas, National and Kapodistrian University of Athens, Greece; Albert Stadler, Herrliberg, Switzerland; David Stone and John Hawkins, Southern Georgia University, Statesboro, GA, and the proposers 5484: Proposed by Mohsen Soltanifar, Dalla Lana School of Public Health, University of Toronto, Canada Let X, X be two continuous positive valued random variables on the real line with corresponding mean, median, and mode,, and,, respectively. Assume for their associated CDFs, (Cumulative Distribution Functions) we have Prove or give a counter eample: F X (t) F X (t) (t > ). (i), (ii), (iii). Solution by Albert Stadler, Herrliberg, Switzerland (i) We have = E(X ) = ( F X (t))dt ( F X (t))dt = E (X ) =. ( ) ( ) (ii) By definition, F X X = F X X =. The functions t F X (t) and t F X (t) are monotonically increasing. Therefore F X ( X ) F X ( X ) implies. (iii) We construct a counter eample as a follows: Let p () =, if / or, p () = 9, if / /, and p () = 9 9, if /.

12 Let p () = 6, if /6, p () = 6, if /6 /, p () =, if or /. p () and p () are probability density functions since Obviously, F X () = p (t)dt Solution by Kee-Wai Lau, Hong Kong, China p ()d = p ()d =. p (t)dt = F X (), however =, = 6. We answer (i) and (ii) in the affirmative and (iii) in the negative. Let X be a continuous random variable on the real line with CDF F X (t) and mean. It is known that = ( F X (t))dt integrals is finite. Since X, X are positive, so (i) follows from the fact that = F X (t)dt, provided at least one of the two (( F X (t)) ( F X (t))) dt = F X (t)dt = F X (t)dt = and (F X (t) F X (t)) dt. Net we consider the medians, assuming that is the least number a satisfying F X (a) =. Suppose, on the contrary, that >. Since F X (t) is a non-decreasing function and < + <, we have F X ( ) = > F X ( + which is false. This proves (ii). ) ) ( + F X F X ( ) =, We now show that (iii) does not necessarily hold. Define the probability density functions f X (t) and f X (t) of X and X as follows:,, 4, 6 f X (t) = and f X (t) = 5, , Then

13 F X (t) =, + 6 4, 4 8, 8, and F X (t) = 5, 5 5. It is easy to check that F X (t) F X (t) for t >, but = 4 < 5 =. This completes the solution. Also solved by the proposer. 5485: Proposed by José Luis Díaz-Barrero, Barcelona Tech, Barcelona, Spain Let, y, z be three positive real numbers. Show that ( + y + z + ) (4 + y + ). cyclic cyclic Solution by Neculai Stanciu, George Emil Palade School, Bazău Romania and Tito Zvonaru, Comănesti, Romania We denote 4 + y + = a, 4y + z + = b, and 4z + + = c. We must prove that ( (a + b)(b + c)(c + a) 8 a + b + ) c ( ) By the AM-GM inequality we have that (a + b)(b + c)(c + a) 8 a + b + c ab bc ca 8 = 8abc 8 = abc, () a b c = abc, () By () and () we obtain (a + b)(b + c)(c + a) 8 ( a + b + ) c abc abc =. I.e. ( ) Solution by Nikos Kalapodis, Patras, Greece By the AM-GM inequality we have ( + y + z + ) (4 + y + ) ( + y + z + ) (4 + y + ). cyclic cyclic cyclic cyclic So, it suffices to prove that ( + y + z + ) (4 + y + ). cyclic cyclic

14 After epanding the inequality reduces to ( +y +z )+ +y +z +(y +yz +z ) ( y+y z+z )+y+yz+z+6yz. Since + y + z y + yz + z, it remains to prove that ( + y + z ) + (y + yz + z ) ( y + y z + z ) + 6yz. This follows again by using the AM-GM inequality properly: ( + y + z ) + (y + yz + z ) = ( + y ) + (y + yz ) + (z + z ) + (y + yz + z ) 4 y + 4y z + 4z + (y + yz + z ) = ( y + y z + z ) + ( y + y z + z + y + yz + z ) ( y + y z + z ) + 6yz. Also solved by Arkady Alt, San Jose, CA; Kee-Wai Lau, Hong Kong, China; Paolo Perfetti, Department of Mathematics, Tor Vergata University, Rome, Italy; Kevin Soto Palacios, Huarmey, Perú; Ioannis D. Sfikas, National and Kapodistrian University of Athens, Greece; Albert Stadler, Herrliberg, Switzerland, and the proposer. 5486: Proposed by Ovidiu Furdui, Technical University of Cluj-Napoca, Cluj-Napoca, Romania Let ( n ) n be the sequence defined by =, =, = and n n+ = n+ + n+ + n + n, n. Prove that the series converges and find n n= its sum. Solution by Ángel Plaza, University of Las Palmas de Gran Canaria, Spain. The recurrence sequence may be unmasked by generating functions. Let F (z) be the associated generating function. That is, F (z) = n z n. Multiplying by z n+ the recurrence relation defining ( n ) and taking into account the initial values it is obtained that F (z) ( z + z ) = z (F (z) z) + z F (z) + z z 4 F (z) + ( z) z( z) + 4 from where F (z) = (z ) ( z z z ). ( Since F (z) converges for z < 7 + ) , then 7 + n = F (/) = 6. n n= Solution by Paolo Perfetti, Department of Mathematics, Tor Vergata University, Rome, Italy Answer: 6. 4

15 Clearly n increases and n. It follows 8 Now we prove the p n= n n = + = p 4 + = p 4 + n= n+ n+ }{{} I n n+ } {{ } I = p p n= + n+ n+ p + n= }{{} I n n = p 4 + n+ n+ = n n+ } {{ } I n } n+ {{} I + n n+ p p+ n= }{{} I p + n= p + n n+ } {{ } I p p n + n+ p p+ p p+ p p+ + n= }{{} I p n= Lemma k / k. Proof of the Lemma n n n+ = p + n n+ = + n+ p p+ p p+ + n= }{{} I p n n+ }{{} /4 as p n n = [ p p+ + p p+ + p p+ + p p+ + p p+ ] + 4 First step: the sequence k / k in monotonic not increasing. that is k+ k+ = k+ + k+ + k + k k+ k+ k+ k+ + k + k k+ k+ k+ and this is implied by (k )+ + (k )+ + (k ) + (k )+ (k )+ + (k )+ + (k ) + (k )+ + (k )+ + k + (k ) = (k )+ via k. The monotonicity of the sequence means that the limit L of k / k does eist and moreover L < +. If L = the proof is concluded yielding lim p 8 p n= n n = 4 n n = 6 L is impossible as shown by the following argument. We employ the Cesaro Stolz theorem that states: k lim k k = lim k+ k k k+ k 5 n=

16 provided that the second limit does eist. We write k+ k+ k+ k+ = k+ + k + k k+ = k The eistence of the limit L = lim would imply k k L = L + 4 L = L = k+ k+ + k 4 k + k k+ Solution by Arkady Alt, San Jose, CA For any sequence ( n ) n let T ( n ) := n+ n+ n+ n, n N {}. Obvious that such defined operator T (we will call it Tribonacci Operator) is linear. ( Since T n ) = n + + n + + n + + n = n then denoting u n := n + n, n N {} we obtain n = u n n, n N {} where T (u n) = and, u =, u = + =, u = + =. Let (t n ) n be the sequence defined by t =, t =, t = and T (t n ) =, n N {}. (Tribonacci Sequence). We have t =, t 4 = 4, t 5 = 7, t 6 =, t 7 = 4, t 8 = 44,... Since det then for any sequence ( n ) n there is triple (c, c, c ) of real 4 numbers such that n = c t n + c t n+ + c t n+, that is sequences (t n ) n, (t n+ ) n, (t n+ ) n form a basis of -dimesion space ker T := {( n ) n T ( n ) =, n N {}}. We will find representation u n as linear combination of t n, t n+, t n+, namely, u n = c t n + c t n+ + c t n+, n N {}. We have u = c t +c t +c t c +c =, u = c t +c t +c t c +c +c =, u = c t + c t + c t 4 c + c + 4c =. From this system of equations we obtain c = c, c c =, c c =.Hence, c =, c =, c = and since u n = t n t n+ + t n+ we obtain n = t n t n+ + t n+ n = t n t n+ + t n+ n. Since radius of convergence of seies n n is and n n = ( ) n= then n n= n = n n= n = = and, therefore, for convergency of ( /) n n= n suffice to prove convergency of series t n n= n. We can prove that using another basis of ker T which form sequences (α n ) n, (β n ) n, (γ n ) n where α, β, γ are roots of characteristic equation =. Substitution = 4u + 4u u = 9 8 n= in equation = give us equivalent equation 6

17 which we solve using substitution u := ( t + ). Then equation 4u u = 9 t 8 ( ( becomes 4 t + )) ( t + ) = 9 t t 8 t + t = 9.Denoting z := t 4 we obtain z + z = 9 9 z =, 9 + t = 9, Since = and u = ( t + ) then suffices to 8 8 t find t = We have t = r (cos ϕ + i sin ϕ), where r = and ϕ = kπ, k =,,. 9 + that is t k = ω k, k =,, and ω = cos π + i sin π, ω =. Thus, denoting θ := 9 +, θ := 9 we obtain α = + θ + θ, β = + ωθ + ω θ, γ = + ω θ + ωθ, the three roots of the equation =. We will prove that α = + θ + θ <. First note that by Power Mean Arithmetic Mean inequality p := < = 9 < 7 = 6. Since = 9 9 = 4 then p = p = 8+ p < 8+ 6 = < 5 = 5. Hence, α <. Also, we obtain β, γ + θ + θ <. Since series ) n, ( β ) n, ( α ( γ n= n= n= of (α n ) n, (β n ) n, (γ n ) n then series k= ) n are convergent and tn is linear combination convergent as well. n= n Now we ready to find sum of series n n= n. Let s n := n t n n and s () = n t n+ n. Note also that function generates k= Tribonacci numbers. Indeed, let a n n ( ) = t n = and since a n n ( ) = a n n a n n+ a n n. Then a n n+ a n n+ = a + (a a ) + (a a a ) + (a n+ a n+ a n+ a n ) n+ then n= a =, a a = a a a = implies a =, a = and 7

18 a n+ a n+ a n+ a n =, n N {}. Thus, a n = t n+, n N {} and, therefore, n t n+ n = s () = k=. In, t n particular, n= n = t n n= n = ( ) s = ( ( ) = 4. ) n Then, n= n = t n n= n t n+ n= n+ + t n+ n= n+ n n= n+ = t n n= n t n n= n + t n n= n n n= n = t + t n n= n = + ( t n n= n t + t ) = ( ) = 6. 4 Solution 4 by Brian D. Beasley, Presbyterian College, Clinton, SC We show that the given series converges by first using induction to prove that n <.95 n for each positive integer n. Note that this claim holds for n {,, }. Given a positive integer k, if n <.95 n for n {k, k +, k + }, then k+ <.95 k k k + k =.95 k (6.755) + k. Thus it suffices to show that.95 k (6.755) + k.95 k+, or equivalently k.95 k (.6675). This latter inequality holds for each positive integer k (using a separate induction argument). Hence n <.95 n for n, so for any positive integer m, m n= n n < n= n n < n=.95 n n = = 9. Since its sequence of partial sums is increasing and bounded above, the given series converges. n Net, we let = L. Then n n= L = n+ + n+ + n + n n+ = 4 + ( L ) + 4 L + 8 L + n n+. Since n n =, we conclude L = 7 8 L () and hence L = 6. Also solved by Kee-Wai Lau, Hong Kong, China; Ioannis D. Sfikas, National and Kapodistrian University of Athens, Greece; Albert Stadler (two solutions), Herrliberg, Switzerland; David Stone and John Hawkins, Southern Georgia University, Statesboro, GA, and the proposer. Mea Culpa 8

19 Arkady Alt of San Jose, CA should have been credited with having solved 5477, and Dionne Bailey, Elsie Campbell, Charles Diminnie, and Trey Smith, all of Angelo State University in San Angelo, TX should have been credited with having solved Paul M. Harms, of North Newton, KS should have been credited for having solved Anna Valkova Tomova of Varna, Bulgaria should have been credited with having solved 5475 and Mea Culpa. 9