已知数列{an},满足an=(2n-1)*2^n,求Sn
2个回答
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an=(2n-1)*2^n
Sn=1*2^1+3*2^2+5*2^3+……+(2n-1)*2^n
2Sn=1*2^2+3*2^3+5*2^4+……+(2n-3)*2^n+(2n-1)*2^(n+1)
Sn-2Sn
=1*2^1+(3-1)*2^2+(5-3)*2^3+……+[(2n-1)-(2n-3)]*2^n-(2n-1)*2^(n+1)
=2+2*(2^2+2^3+……+2^n)-(2n-1)*2^(n+1)
=2+2*4*[1-2^(n-1)]/(1-2)-(2n-1)*2^(n+1)
=2+8*2^(n-1)-8-(2n-1)*2^(n+1)
=-(2n-3)*2^(n+1)-6
所以Sn=(2n-3)*2^(n+1)+6
Sn=1*2^1+3*2^2+5*2^3+……+(2n-1)*2^n
2Sn=1*2^2+3*2^3+5*2^4+……+(2n-3)*2^n+(2n-1)*2^(n+1)
Sn-2Sn
=1*2^1+(3-1)*2^2+(5-3)*2^3+……+[(2n-1)-(2n-3)]*2^n-(2n-1)*2^(n+1)
=2+2*(2^2+2^3+……+2^n)-(2n-1)*2^(n+1)
=2+2*4*[1-2^(n-1)]/(1-2)-(2n-1)*2^(n+1)
=2+8*2^(n-1)-8-(2n-1)*2^(n+1)
=-(2n-3)*2^(n+1)-6
所以Sn=(2n-3)*2^(n+1)+6
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