Question

已知数列{an}的通项公式为an=1/n(n+2）求前n项的和

1，已知数列{an}的通项公式为an=1/n(n+2）求前n项的和
2，求Sn=1+9+45+..........+(2n)3^n-1
3,Sn=1+2乘2+3乘2^2+...........n乘2^(n-1)
数列{an}中若a1+a2+a3+......+an=(n^2)+1的通项公式

Answer 1

1 an=1/[n(n+2)]
Sn=1/(1×3)+1/(2×4)+1/(3×5)+1/(4×6)+···1/[(n-1)(n+1)]+1/[n(n+2)]
=(1/2)[(1-1/3)+(1/2-1/4)+(1/3-1/5)···(1/n-1/(n+2))]
=(1/2)[1+1/2-1/(n+1)-1/(n+2)]
这种方法叫做裂项相消 然后化简一下就行了
2 3两题就是要错位相减，适用于一个等差和一个等比相乘情况下的求和
第二题好像有问题 通项是错的
3
Sn=(1×2^0)+(2×2^1)+(3×2^2)+···+[n×2^(n-1)](1)
2Sn= (1×2^1)+(2×2^2)···+[(n-1)×2^(n-1)]+[n×2^n](2)
（1）-(2)得
-Sn=1-n×2^n+[2^1+2^2+···+2^(n-1)]
Sn=1+(n+1)2^n

Answer 2

第一题:Sn = 1/2 ×(1/n - 1/(n+2) + 1/(n-1) - 1/(n+1) + 1/(n-2) - 1/n + ....+ 1/4 -1/6 + 1/3 - 1/5 + 1/2 - 1/4 + 1 - 1/3) = 3/4 - 1/2(n+2) - 1/2(n+1)

最后的补充题:an = Sn - S(n-1) = (n^2)+1 - ((n-1)^2)+1 = 2n + 1

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