Question

cn=[2∧(n+2)－2]/(n+1)(n+2)+[2∧(n+1)]n/n+1,求cn前n项和

Answer 1

根据题目所给的公式，第n项为：
cn = (2^(n+2) - 2) / [(n+1) * (n+2)] + (2^(n+1) * n) / (n+1)
则前n项的和为：
c1 + c2 + ... + cn
= [(2^3 - 2) / (2 * 3) + (2^2 * 1) / 2]
+ [(2^4 - 2) / (3 * 4) + (2^3 * 2) / 3]
+ [(2^5 - 2) / (4 * 5) + (2^4 * 3) / 4]
+ ...
+ [(2^(n+2) - 2) / [(n+1) * (n+2)] + (2^(n+1) * n) / (n+1)]
= [6(2-2/2) + 2(4-2)/2] / 2 + [12(2-2/3) + 8(3-1)/3] / 6 + [20(2-2/4) + 16(4-1)/4] / 12 + ... + [2^(n+2)(n-1)/(n+1)(n+2) + 2^(n+1)n/(n+1)]
= 1/2{[6 - 2/1] + [12 - 2/2] / 2 + [20 - 2/3] / 3 + ... + [2^(n+1) - 2/(n-1)] / (n-1) + [2^n * n] / (n+1)}
= 1/2(4 + 5/2 + 14/3 + ... + [2^(n+1) - 2/(n-1)] / (n-1) + [2^n * n] / (n+1))
注：这里利用了等比数列求和公式和小学奥数的魔术方法。
因此，前n项的和为 1/2(4 + 5/2 + 14/3 + ... + [2^(n+1) - 2/(n-1)] / (n-1) + [2^n * n] / (n+1))。