数列{an}的通项公式an=(2n+2)2的n次方-1,求Sn
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an=(2n+2)*2^n-1
an=(n+1)*2^(n+1)-1
sn=2*2^2-1+3*2^3-1+.+(n+1)*2^(n+1)-1
sn=2*2^2+3*2^3+.+(n+1)*2^(n+1)-n
2sn=2*2^3+3*2^4+.+n*2^(n+1)+(n+1)*2^(n+2)-2n
sn-2sn=2*2^2+2^3+2^4+.+2^(n+1)-(n+1)*2^(n+2)+2n
-sn=2*2^2+8*[1-2^(n-1)]/(1-2)-(n+1)*2^(n+2)+2n
-sn=8+8*2^(n-1)-8-(n+1)*2^(n+2)+2n
-sn=2^(n+2)-(n+1)*2^(n+2)+2n
sn=(n+1)*2^(n+2)-2^(n+2)-2n
sn=n*2^(n+2)-2n
an=(n+1)*2^(n+1)-1
sn=2*2^2-1+3*2^3-1+.+(n+1)*2^(n+1)-1
sn=2*2^2+3*2^3+.+(n+1)*2^(n+1)-n
2sn=2*2^3+3*2^4+.+n*2^(n+1)+(n+1)*2^(n+2)-2n
sn-2sn=2*2^2+2^3+2^4+.+2^(n+1)-(n+1)*2^(n+2)+2n
-sn=2*2^2+8*[1-2^(n-1)]/(1-2)-(n+1)*2^(n+2)+2n
-sn=8+8*2^(n-1)-8-(n+1)*2^(n+2)+2n
-sn=2^(n+2)-(n+1)*2^(n+2)+2n
sn=(n+1)*2^(n+2)-2^(n+2)-2n
sn=n*2^(n+2)-2n
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