作业帮求数列an/bn=(2n-1)/(2^(n-1))的前n项和Sn?
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求数列an/bn=(2n-1)/(2^(n-1))的前n项和Sn?
求数列an/bn=(2n-1)/(2^(n-1))的前n项和Sn?
求数列an/bn=(2n-1)/(2^(n-1))的前n项和Sn?
Sn=1/2^0+3/2^1+5/2^2+……+(2n-1)/(2^(n-1))
2Sn=1/2^(-1)+3/2^0+5/2^1+……+(2n-1)/(2^(n-2))
2Sn-Sn=2+2/2^0+2/2^1+2/2^2+……+2/(2^(n-2))-(2n-1)/(2^(n-1))
=2+2(1-(1/2)^(n-1))/(1-1/2)-(2n-1)/(2^(n-1))
=6-(2n+3)/2^(n-1)