數學問題:SEQUENCE ! 謝謝大大。

tat

2011-10-04 7:46 am

sequence 兩條:
第一條: Find the sum of the series 1xn + 2x(n-1)+.....+ nx1.
第二條: Prove that the sum of the first n terms of the series
1^2 + 2x2^2 +3^2 + 2x4^2 +5 ^2 + 2x6^2 +... is 1/2n(n+1)^2 when N is even.
Find a similar formula for the sum when n is odd.
唔該哂各位師兄師姐=) ...

回答 (1)

myisland8132

2011-10-04 8:03 am

✔ 最佳答案

1 1xn + 2x(n-1)+.....+ nx1

= Σ k(n + 1 - k)

= nΣ k + Σ k - Σ k^2

= n^2(n + 1)/2 + n(n + 1)/2 - n(n + 1)(2n + 1)/6

= n(n + 1)^2/2 - n(n + 1)(2n + 1)/6

= n(n + 1)/6 * [3(n + 1) - (2n + 1)]

= n(n + 1)(n + 2)/6

2 when n is even

1^2 + 2x2^2 +3^2 + 2x4^2 +5 ^2 + 2x6^2 +... 2xn^2

= (1^2 + 2^2 + ... + n^2) + (2^2 + 4^2 + ... + n^2)

= n(n + 1)(2n + 1)/6 + 4(1^2 + 2^2 + ... + (n/2)^2)

= n(n + 1)(2n + 1)/6 + 4[(n/2)(n/2 + 1)(n + 1)/6]

= n(n + 1)(2n + 1)/6 + n(n + 2)(n + 1)/6

= n(n + 1)^2/2

when n is odd

1^2 + 2x2^2 +3^2 + 2x4^2 +5 ^2 + 2x6^2 +... n^2

= (1^2 + 2^2 + ... + n^2) + (2^2 + 4^2 + ... + (n - 1)^2)

= n(n + 1)(2n + 1)/6 + 4(1^2 + 2^2 + ... + (n - 1)/2)^2)

= n(n + 1)(2n + 1)/6 + 4[(n - 1)/2)((n - 1)/2 + 1)(n)/6]

= n(n + 1)(2n + 1)/6 + n(n - 1)(n + 1)/6

= n^2(n + 1)/2

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