Cn=anbn,求数列{an}的前n项和,bn=2n+1,an=2^n_数学解答

Cn=anbn,求数列{an}的前n项和,bn=2n+1,an=2^n

人气：131 ℃　时间：2019-10-08 10:40:52

解答

cn=anbn=(2n+1)×2ⁿ=n×2^(n+1)+2ⁿ
Sn=c1+c2+...+cn=[1×2²+2×2³+...+n×2^(n+1)] +(2+2²+...+2ⁿ)
令Cn=1×2²+2×2³+...+n×2^(n+1)
则2Cn=1×2³+2×2⁴+...+(n-1)×2^(n+1)+n×2^(n+2)
Cn-2Cn=-Cn=2²+2³+...+2^(n+1)-n×2^(n+2)
Cn=n×2^(n+2)-[2²+2³+...+2^(n+1)]
Sn=Cn+(2+2²+...+2ⁿ)
=n×2^(n+2)-[2²+2³+...+2^(n+1)]+(2+2²+...+2ⁿ)
=n×2^(n+2)+2 -2^(n+1)
=(2n-1)×2^(n+1) +2