Question

急！an=n(n+1)/2^n，求Sn？

给点思路也行啊，很急啊！还可以加分的！
n(n+1)
an= __________ ，求Sn？
2^n

Answer 1

解：因为an=n(n+1)/2^n
所以a1=1*2/2,a2=2*3/2^2,a3=3*4/2^3……
因此Sn=a1+a2+a3……+an
=1*2/2+2*3/2^2+3*4/2^3+……n(n+1)/2^n ____(1)
1/2Sn=1*2/2^2+2*3/2^3+3*4/2^4+……n(n+1)/2^(n+1)____(2)
(1)-(2)斜相消得
1/2Sn=1+2/2+3/4+4/8+5/16+……n/2^(n-1)-n(n+1)/2^(n+1)____(3)
1/4Sn=1/2+2/4+3/8+4/16+5/32+……n/2^(n)-n(n+1)/2^(n+2)____(4)
(3)-(4)斜相消得
1/4Sn=1+1/2+1/4+1/8+1/16+1/32+……+1/2^(n-1)-n/2^(n)-n(n+1)/2^(n+2)
=2-1/2^(n-1)-n/2^(n)-n(n+1)/2^(n+2)
=2-1/2^(n-1)-n(n+5)/2^(n+2)
所以Sn=8-1/2^(n-3)-n(n+5)/2^n
代入检验，全都符合

Answer 2

a(n+1)=(n+1)(n+2)/2^(n+1)=(1/2)an +(n+1)/2^n=(1/2)an +(n/2^n) +(1/2^n)

设：cn=1/2^n
这个数列的前n项和Qn=(1/2)[1-(1/2)^n]/[1-(1/2)]=1-(1/2^n)

设：bn=n/2^n, 这个数列的前n项和Pn
b(n+1)=(n+1)/2^(n+1)=(1/2)bn +(1/2)(1/2^n)
2b(n+1)=bn+(1/2^n)=bn+cn
2P(n+1)-2b1=Pn+Qn
b(n+1)+P(n+1)-2b1=Qn
P(n+1)=Qn-b(n+1)+2b1=2-(1/2^n)-b(n+1)
Pn=P(n+1)-b(n+1)=2-(1/2^n)-2b(n+1)=2-(1/2^n)-2(n+1)/2^(n+1)=2-(1/2^(n-1))-(n/2^n)

由a(n+1)=(1/2)an +(n/2^n) +(1/2^n)
2a(n+1)=an+2bn+2cn
2S(n+1)-2a1=Sn+2Pn+2Qn
a(n+1)+S(n+1)-2a1=2Pn+2Qn
S(n+1)=2a1+2Pn+2Qn-a(n+1)
Sn=S(n+1)-a(n+1)=2a1+2Pn+2Qn-2a(n+1)=2+2Pn+2Qn-[(n+1)(n+2)/2^n]
将Pn,Qn代入，整理一下就可以了。

Answer 3

图片可以放大了查看。

Answer 4

an=n(n+1)/2^n 2an=n(n+1)/2^(n-1)
an-1=n(n-1)/2^(n-1) 2an-1=n(n-1)/2^(n-2)
..
a2=6/2^2 2a2=6/2
a1=1 2a1=2
2Sn-(Sn-(n(n+1)/2^n)=2+[2+3/2+..+n/2^(n-2)]
Sn=2+[2+3/2+..+n/2^(n-2)]-(n(n+1)/2^n

设bn=n/2^(n-2) 2bn=n/2^(n-3)
bn-1=(n-1)/2^(n-3) 2bn-1=(n-1)/2^(n-4)
..
b2=2/1 2b2=4/1
2(b2+..+bn)-(b2+..+bn-1)=1+1/2+..+1/2^(n-3)=1(1-1/2^(n-2))/(1-1/2)=2(1-1/2^(n-2)
b2+..+2bn=2(1-1/2^(n-2))
b2+..+bn=2(1-1/2^(n-2))-n/2^(n-2)

Sn=2+2(1-1/2^(n-2))-n/2^(n-2)-n(n+1)/2^n

Answer 5

令bn(x)=[(x)^(n+1)]
Tn(x)=x/(1-x) an(x)=n(n+1)x^=b''n*x
sn(x)=xT''n(x)=2x/(1-x)^3
an=an(1/2)=8