an=(2n+1)*2^n,求Sn

gier2022-10-04 11:39:541条回答

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jasmine0502 共回答了20个问题 | 采纳率75%: 错位相减
a1=3*2^1
a2=5*2^2
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an=(2n+1)*2^n
Sn=3*2^1+5*2^2+.+(2n+1)*2^n
2Sn= 3*2^2+5*2^3+.+(2n-1)*2^n+(2n+1)*2^(n+1)
2Sn-Sn=-3*2^1-2*2^2-2*2^3-.-2*2^n+(2n+1)*2^(n+1)
Sn=-2^1-2*2^1-2*2^2-2*2^3-.-2*2^n+(2n+1)*2^(n+1)
Sn=-2(1+2^1+2^2+2^3+.+2^n)+(2n+1)*2^(n+1)
Sn=-2*1*(1-2^(n+1))/(1-2)+(2n+1)*2^(n+1)
Sn=2*(1-2^(n+1))+(2n+1)*2^(n+1)
Sn=2-2*2^(n+1)+(2n+1)*2^(n+1)
Sn=2+(2n-1)*2^(n+1); 1年前