An=n*(2^n-1),求Sn
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解:sn=1*2^1+2*2^2+3*2^3+...+n*2^n-n(n+1)/2(1)
(1)式两边同乘以2得
2sn=1*2^2+2*2^3+3*2^4+...+(n-1)*2^n+n*2^(n+1)-n(n+1) (2)
(2)减去(1)得sn=1*2^1-2^2-2^3-2^4-2^n+n*2^(n+1)-n(n+1)/2
=2-(2^2+2^3+...+2^n)+n*2^(n+1)-n(n+1)/2
=2-{2^2[2^(n-1)-1]+n*2^(n+1)-n(n+1)/2
=6-2^(n+1)+n*2^(n+1)-n(n+1)/2
(1)式两边同乘以2得
2sn=1*2^2+2*2^3+3*2^4+...+(n-1)*2^n+n*2^(n+1)-n(n+1) (2)
(2)减去(1)得sn=1*2^1-2^2-2^3-2^4-2^n+n*2^(n+1)-n(n+1)/2
=2-(2^2+2^3+...+2^n)+n*2^(n+1)-n(n+1)/2
=2-{2^2[2^(n-1)-1]+n*2^(n+1)-n(n+1)/2
=6-2^(n+1)+n*2^(n+1)-n(n+1)/2
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