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数列{an}的通项公式an=(2n+2)乘2的n-1次方求Sn

题目详情
数列{an}的通项公式an=(2n+2)乘2的n-1次方
求Sn
▼优质解答
答案和解析
由a(n)=(2n+2)×2^(n-1)=(n+1)×2^n=n×2^n+2^n,得:
a(1)=1×2^1+2^1
a(2)=2×2^2+2^2
a(3)=3×2^3+2^3
······
a(n)=n×2^n+2^n
将上述n个式子相加,得:
S(n)=(1×2^1+2×2^2+3×2^3+······+n×2^n)+(2^1+2^2+2^3+······+2^n).
令M=1×2^1+2×2^2+3×2^3+······+n×2^n,
则2M=1×2^2+2×2^3+3×2^4+······+(n-1)×2^n+n×2^(n+1)
∴M-2M=2^1+(2-1)×2^2+(3-2)×2^3+······+[n-(n-1)]×2^n-n×2^(n+1)
∴M=n×2^(n+1)-(2^1+2^2+2^3+······+2^n).
∴S(n)=M+(2^1+2^2+2^3+······+2^n)=n×2^(n+1).