Question

求数列｛n/2^n｝的前n项和Sn

Answer 1

sn=1*×(2^(1-1))
2*×(2^(2-1))
3*×(2^(3-1))
...
n×(2^(n-1))=1*2^0
2*2^1
3*2^2
...
n×(2^(n-1))
2sn=[1*×(2^(1-1))
2*×(2^(2-1))
3*×(2^(3-1))
...
n×(2^(n-1))]*2=1*2^1
2*2^2
3*2^3
...
n×(2^n)
sn=2sn-sn=[1*2^1
2*2^2
3*2^3
...
n×2^n]-[1*2^0
2*2^1
3*2^2
...
n×(2^(n-1))]
=n×(2^n)-1*2^0
(1*2^1-2*2^1)
(2*2^2-3*2^2)
...
[(n-1)*×(2^(n-1))-n×(2^(n-1))]
=n×(2^n)-1*2^0
[-2^1-2^2-....-2^(n-1)]=n×(2^n)-1*2^0-[2^1
2^2
....
2^(n-1)]
因为2^1
2^2
....
2^(n-1)为首相是2^1
公比为2的等比数列
所以和为【2^1-2^(n-1)】/(1-2)
所以sn=n×(2^n)-1*2^0-[2^1
2^2
....
2^(n-1)]=n×(2^n)-1*2^0-【2^1-2^(n-1)】/(1-2)=n×(2^n)-2^(n-1)
1
望你能满意

Answer 2

解：
Sn=
(1/2^1)+(2/2^2)+(3/2^3)+…………
+[(n-1)/2^(n-1)]+(n/2^n),
则2Sn=(1/2^0)+(2/2^1)+(3/2^2)+……+[(n-1)/2^(n-2)]+[n/2^(n-1)],
两式相减得到：
Sn=2Sn-Sn
=(1/2^0)+
{
(1/2^1)+(1/2^2)+(1/2^3)+……+[1/2^(n-1)]
}
-(n/2^n),
【{}中为首项为1/2，公比为1/2的等比数列前(n-1)项和】
=1-(n/2^n)
+
(1/2)[1-(1/2)^(n-1)/[1-(1/2)]
=1-(n/2^n)
+(1-2/2^n)
=2-
(n+2)/2^n.
即Sn=2-
(n+2)/2^n.