an=(2n+1)*2^n,求Sn

来源：学生作业帮助网编辑：作业帮时间：2024/11/26 23:31:50

x��)�K̳�0��6��2��y��)8�&�H��~�� n/��|��O��!F��B8F��@��iX$8�C�T[OM��@�,RaQ�Q�P��at�zt��t�Ic]=0�UHL9i�4�A��:�bM��4��C�5�Al\��U�RvX� DhA��@� Ɖ�

an=(2n+1)*2^n,求Sn
an=(2n+1)*2^n,求Sn

an=(2n+1)*2^n,求Sn
错位相减
a1=3*2^1
a2=5*2^2
.
an=(2n+1)*2^n
Sn=3*2^1+5*2^2+.+(2n+1)*2^n
2Sn= 3*2^2+5*2^3+.+(2n-1)*2^n+(2n+1)*2^(n+1)
2Sn-Sn=-3*2^1-2*2^2-2*2^3-.-2*2^n+(2n+1)*2^(n+1)
Sn=-2^1-2*2^1-2*2^2-2*2^3-.-2*2^n+(2n+1)*2^(n+1)
Sn=-2(1+2^1+2^2+2^3+.+2^n)+(2n+1)*2^(n+1)
Sn=-2*1*(1-2^(n+1))/(1-2)+(2n+1)*2^(n+1)
Sn=2*(1-2^(n+1))+(2n+1)*2^(n+1)
Sn=2-2*2^(n+1)+(2n+1)*2^(n+1)
Sn=2+(2n-1)*2^(n+1)

an=n*2^n,求Sn an=2^n+n,求Sn a1=1/2,Sn=n^2an-n(n-1),求Sn和an an=(2^n+1/(2^n))^2求Sn an=(2n-1)*2^(n-1),求Sn an=(2n-1)2^n-1,求sn? an=(2n+1)*2^n,求Sn 已知an=1/2n(n+1),求Sn 知Sn=n²+2n+1求an 已知an=(2n+1)*3^n,求Sn An=n×2^(n-1),求Sn an=（2^n-1）n,求Sn 已知:Sn=n平方+2n+1,求An a1=1,Sn=n^2an,求an Sn=2an+(-1)n次方怎么求an sn=2*3^n,求an 求an的前n项和Sn an=(2n-1)*2^(n-1) an=2^n+(2n-1)求an的前n项和sn