Question

已知数列的通项公式为an=2n-1

已知数列an的通项公式an＝2n-1,数列bn的通项公式为bn＝1/2^n,求数列an*bn的前n项和Sn

Answer 1

an为等差数列,bn为等比数列,求和方法错位相减法
Sn=1/2+3*1/4+5*1/8+7*1/16+.+(2n-1)*1/2^n ①
1/2Sn=1/4+3*1/8+5*1/16+.+(2n-3)*1/2^n+(2n-1)*1/2^(n+1) ②
①-②:
1/2Sn=1/2+2(1/4+1/8+1/16+.+1/2^n)-(2n-1)/2^(n+1)
=1/2+2*[1-1/2^(n-1)]/(1-1/2)-(2n-1)/2^(n+1)
=1/2+4-(2n+15)/2^(n+1)
=9/2-(2n+15)/2^(n+1)
Sn=9-(2n+15)/2^n
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