A difficult question

用戶85377

2009-11-10 3:24 am

Given that [1/(n + 2)] (1+ x)^(n+2) = (n+1)C0 x + 1/2 (n+1)C1 x^2 + 1/3(n+1)C2 x^3 + .... + 1/(n + 2) (n+1)C(n+1) x^ (n + 2) + 1/(n+2)

Find, in terms of n,
nC0 + 5/2 (n + 1)C1 + 5^2/3 (n + 1)C2 +....+ {[5^(n+1)]/(n + 2) } (n+1)C(n+1)

更新1:

There are some amendment. Just forgot to add the bracket. (5/2) (n + 1)C1 + [(5^2)/3] (n + 1)C2

回答 (1)

用戶9112

2009-11-10 4:04 am

✔ 最佳答案

[1/(n + 2)] (1+ x)^(n+2) = (n+1)C0 x + 1/2 (n+1)C1 x^2 + 1/3(n+1)C2 x^3 + .... + 1/(n + 2) (n+1)C(n+1) x^ (n + 2) + 1/(n+2)
Put x = 5,
[1/(n + 2)] (1+ 5)^(n+2) = (n+1)C0 (5) + 1/2 (n+1)C1(5)^2 + 1/3(n+1)C2 (5)^3 + .... + 1/(n + 2) (n+1)C(n+1) (5)^(n + 2) + 1/(n+2)
[6^(n+2)] / (n + 2) - 1/(n + 2) = (n+1)C0 (5) + 1/2 (n+1)C1(5)^2 + 1/3(n+1)C2 (5)^3 + .... + 1/(n + 2) (n+1)C(n+1) (5)^(n + 2)
{[6^(n + 2)] - 1} / (n + 2) = (n+1)C0 (5) + 1/2 (n+1)C1(5)^2 + 1/3(n+1)C2 (5)^3 + .... + 1/(n + 2) (n+1)C(n+1) (5)^(n + 2)
{[6^(n + 2)] - 1} / [5(n + 2)] = (n+1)C0 + (5/2)(n+1)C1 + [(5^2)/3](n+1)C2 + .... + [5^(n+1)]/(n + 2)} (n+1)C(n+1)
{[6^(n + 2)] - 1} / [5(n + 2)] = nC0 + (5/2)(n+1)C1 + [(5^2)/3](n+1)C2 + .... + [5^(n+1)]/(n + 2)} (n+1)C(n+1) since nC0 = (n+1)C0 = 1
The required answer is {[6^(n + 2)] - 1} / [5(n + 2)]

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