Mathematical induction

2008-08-14 1:43 am

Given that 1x5 + 2x6 + 3x7 + ... + n(n+4) = [n(n+1)(2n+13)] / 6
and 1+2+3+...+n=[n(n+1)]/2

Find the sum of 1^2 + 2^2 +3^2 +....+ n^2 .

回答 (1)

Kestrel

2008-08-14 2:07 am

✔ 最佳答案

1x5 + 2x6 + 3x7 + ... + n(n+4)
= 1 x ( 1 + 4 ) + 2 x ( 2 + 4 ) + 3 x ( 3 + 4 ) + ... + n ( n + 4 )
= 1 x 1 + 1 x 4 + 2 x 2 + 2 x 4 + 3 x 3 + 3 x 4 + ... + n x n + n x 4
= ( 1 x 1 + 2 x 2 + 3 x 3 + ... + n x n ) + ( 1 x 4 + 2 x 4 + 3 x 4 + ... + n x 4 )
= ( 1^2 + 2^2 +3^2 +....+ n^2 ) + 4 ( 1+2+3+...+n )

即,
( 1^2 + 2^2 +3^2 +....+ n^2 ) + 4 ( 1+2+3+...+n ) = 1x5 + 2x6 + 3x7 + ... + n(n+4)
所以,
1^2 + 2^2 +3^2 +....+ n^2 = [1x5 + 2x6 + 3x7 + ... + n(n+4)] - [ 4 ( 1+2+3+...+n )]
1^2 + 2^2 +3^2 +....+ n^2 = [n(n+1)(2n+13)]/6 - 4 x [n(n+1)]/2
= [n(n+1)(2n+1)]/6.

參考: ME

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