Question

高二数学 已知an=(n+2)/2^n,求sn 求详解!

Answer 1

分析，
an=(n+2)/2^n
=n/2^n+1/2^(n-1)【分成两部分计算】

Sn=a1+a2+……+an
=(1/2+2/2²+3/2³……+n/2^n)+(1+1/2+……+1/2^(n-1))
设tn=(1/2+2/2²+3/2³……+n/2^n)
2tn=1+2/2+3/2²+4/2³+……+n/2^(n-1)
tn=2tn-tn【错项相减】
=1+1/2+1/2²+……+1/2^(n-1)-n/2^n
=2-2^(1-n)-n/2^n
∴Sn=tn+2-2^(1-n)
=4-(4+n)/2^n

Answer 2

Sn= 3/2+4/2^2+5/2^3+…… +(n+1)/2^(n-1)+(n+2)/2^n ……①
2Sn=3+4/2+5/2^2+6/2^3+……+(n+2)/2^(n-1) ……②
②-①,得：
Sn=3+(4-3)/2+(5-4)/2^2+(6-5)/2^3+……+(n+2-(n+1))/2^(n-1)-(n+2)/2^n
=2+1+1/2+1/2^2+……+1/2^(n-1)-(n+2)/2^n
=2+(1-(1/2)^n)/(1-1/2)-(n+2)/2^n
=4-2/2^n-(n+2)/2^n
=4-(n+4)/2^n