Examples on Mathematical Induction: Sum of product of integers Created by Mr. Francis Hung Last updated: December 4, 2017
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1 Examples o Mathematical Iductio: Sum of product of itegers Created by Mr. Fracis Hug Last updated: December, 07. Prove that = for =,,, Let P() = for =,,, =, LHS = = ; RHS = = P() is true. Suppose P() is true. i.e = for some > 0 Whe = +, LHS = ( + ) + = + ( + ) + = = = = RHS If P() is true the P( + ) is also true. By the priciple of mathematical iductio, P() is true for all positive.. 00 Additioal Mathematics Q (a) (b) Prove, by mathematical iductio, that () + ( ) + ( ) + + ( + )( ) = ( + ) for all positive itegers. Show that () + ( ) + ( ) ( 98 ) = 97( 99 ) + (a) Whe =, LHS = () =, RHS = () =, LHS = RHS The statemet is true for =. Suppose () + ( ) +( ) + + ( + )( ) = ( + ) for some positive iteger. Whe = +, LHS = () + ( ) +( ) + + ( + )( ) + ( + )( + ) = ( + ) + ( + )( + ) = ( + )( + ) = ( + )( + ) = RHS The statemet is also true for = +. By the priciple of mathematical iductio, it is true for all positive iteger. (b) () + ( )+( )+ +98( 98 ) = () + ( )+( )+ +99( 98 ) [ ] = 98( ) = 98( 99 ) 99 + = 97( 99 ) +. Prove that ( + ) = Hece fid the value of ad + ( + ) + ( + + ) + + ( ). for all positive iteger. Page
2 Sum of product of itegers Mr. Fracis Hug Page. Prove that (+) = 7 for all positive iteger.. Prove that () = ( + )( + ) for all positive iteger.. Prove that = ( + )( + ) for all positive iteger Q Prove that ( + ) = for all positive itegers. Let P() ( + ) =, where is a positive iteger. =, LHS = =, RHS = = LHS = RHS P() is true. Suppose ( + ) = for some positive iteger. Whe = +, LHS = ( + ) + ( + )( + ) = + ( + )( + ) = = 8 = = If P() is true the P( +) is also true. By M.I., P() is true for all positive iteger. 8. Prove that +( )+( )++( )+= ( + )( + ) for all positive itegers. Let P() =. LHS = ; RHS = = P() is true. Suppose P() is true. = = = If P() is true, the P( + ) is also true. By the priciple of mathematical iductio, P() is true for all atural umbers.
3 Sum of product of itegers Mr. Fracis Hug for all positive itegers. Hece simplify ( ) + ( ) + ( ) [ ( )]. (a) Let P() ( ) = for =,,,... =, LHS =, RHS = It is true for = Suppose ( ) = Whe = +, LHS = ( ) + ( + )( + ) = + ( + )( + ) (iductio assumptio) = 7 (= ) = = = = = RHS If P() is true, the P( + ) is also true. By the priciple of mathematical iductio, it is true for all positive itegers. (b) ( ) + ( ) + ( ) [ ( )] = [ ( )] = ( ) = = = 9. Prove that ( ) = Page
4 Sum of product of itegers Mr. Fracis Hug 0. Prove, by mathematical iductio, that ( + )( ) = 9 Let P() ( + )( ) = 9 =, L.H.S. = =, R.H.S. = 9 = for all positive itegers., is a +ve iteger. L.H.S. = R.H.S. P() is true. Suppose that P() is true for some positive iteger. i.e ( + )( ) = 9 (*) Whe = +, L.H.S. = ( + )( ) + ( + )[( + ) ] = 9 + ( + )( + ) (by (*)) = 9 = 9 0 = 9 R.H.S.= 9 = = 7 = 7 7 = 9 If P() is true the P( + ) is also true By the priciple of mathematical iductio, P() is true for all positive iteger.. 99 Paper Q Prove by mathematical iductio, that ( + ) = for ay positive iteger. Page
5 Sum of product of itegers Mr. Fracis Hug. 998 Paper Q Prove, by mathematical iductio, that ( + ) = () for all positive itegers. Let P() ( + ) = (), is a positive iteger. =, L.H.S. = =, R.H.S. = () = L.H.S. = R.H.S. P() is true. Suppose that P() is true for some positive iteger. i.e ( + ) = () (*) Whe = +, L.H.S. = ( + )+ ( + ) = () + ( + ) (by (*)) = ( + + ) = ( + ) = + ( + ) = R.H.S. If P() is true the P( + ) is also true By the priciple of mathematical iductio, P() is true for all positive iteger.. 99 Paper Q 0 M Q Prove by mathematical iductio, that ( ) = ( + ) for all positive itegers. Let P() ( ) = ( + ), where is a positive iteger. =, L.H.S. = =, R.H.S. = ( + ) = L.H.S. = R.H.S. P() is true. Suppose that P() is true for some positive iteger. i.e ( ) = ( + ) (*) Whe = +, L.H.S. = ( ) + ( + )[( + ) ] = ( + ) + ( + )( + ) (by (*)) = ( + )( + + ) = ( + ) ( + ) = R.H.S. If P() is true the P( + ) is also true By the priciple of mathematical iductio, P() is true for all positive iteger. Page
6 Sum of product of itegers Mr. Fracis Hug. (a) Prove, by mathematical iductio, that for all positive itegers, ( + ) = ( + )( + + ). (b) Hece, accordig to the formula = ( + )( + ), deduce the formula for evaluatig ( + ). (a) Let P() ( + ) = ( + )( + + ). =, LHS = =, RHS = ( + )( + + ) = LHS = RHS P() is true. Suppose P() is true for some positive iteger. i.e ( + ) = ( + )( + + ) Add ( + ) [( + ) + ] to both sides. LHS = ( + ) + ( + ) [( + ) + ] = ( + )( + + ) + ( + ) ( + ) (iductio assumptio) = ( + )( + + ) + ( + )( + + ) = ( + )( ) = ( + )( ) RHS = ( + )( + + )[( + ) + ( + ) + ] = ( + )( + )( ) = ( + )( ) = ( + )( ) LHS = RHS If P() is true the P( + ) is also true. By the priciple of mathematical iductio, the formula is true for all positive iteger. (b) ( + ) = ( ) + ( ) + ( ) + + ( + ) = ( + ) ( ) = ( + )( + + ) ( + )( + ) = ( + )( ) = ( + )( ) = ( + )( + )( + ) Page
7 Sum of product of itegers Mr. Fracis Hug. Prove that (+)(+)(+7) = 7 0 for all positive itegers. Let P() (+)(+)(+7)= =, LHS = 70, RHS = 70 70= LHS = RHS, P() is true. Suppose (+)(+)(+7)= Whe = +, LHS = (+)(+)(+7)+( + )( + 7)( + 0) = (+)(+7)(+0) = = = If P() is true, the P( + ) is also true. By the priciple of mathematical iductio, it is true for all positive itegers.. (a) Prove that... ( )( ) ( )( )( ) for all positive iteger. (b) Hece, or otherwise, fid the value of (a) Let P() be the statemet. Whe =, LHS = =, RHS = ( )( )( ). P() is true. Assume P() is true. i.e. +++( + )( + ) = ( + )( + )( + ) Whe = +, LHS = +++( + )( + ) + ( +)( + )( + ) = ( )( )( ) ( )( )( ) = ( )( )( )( ) = ( )( )( )( ) = ( )( )( )( ) = RHS If P() is true the P(+) is also true. By the Priciple of M.I., the statemet is true for all positive itegers. (b)... 7 = (... 7) (... 0 ) = ( )( )( ) 0(0 )(0 )(0 ) = = 80 Page 7
8 Sum of product of itegers Mr. Fracis Hug for all positive itegers Paper Q7 Let T = ( + )(!) for ay positive iteger. Prove, by mathematical iductio, that T + T + + T = [( + )!] for ay positive iteger. [Note:! = ( )( ).] 9. Let P() ()(!) + ()(!) + (!) + + ()(!) = ( + )!. 7. Prove that to th term = =, LHS = ()(!) =, RHS =! = P() is true. Suppose ()(!) + ()(!) + (!) + + ()(!) = ( + )! for some positive iteger. Add ( + )( + )! To both sides LHS = ()(!) + ()(!) + (!) + + ()(!) + ( + )( + )! = ( + )! + ( + )( + )! (iductio assumptio) = ( + )! ( + + ) = ( + )! ( + ) = ( + )! = RHS If P() is true the P( + ) is also true. By the priciple of mathematical iductio, P() is true for all positive itegers. p x x x x, prove that U xu p x.! 0. Let U (x) = p Q(p) U xu x p =, LHS = U p x = U (x) = x x RHS = U x = = x! Q() is true. Suppose U x U x U Page 8, for all positive iteger p. for some positive iteger. x U xu x = U x U x x x x xx x x x =! x x! x x x xx x =! x x x x =! = U x If Q() is true, the Q( + ) is also true. By the priciple of mathematical iductio, it is true for all positive itegers.
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