Question

数列{n·2^n-1}的前n项和Sn=

Answer 1

Sn=2^n-1

S(n-1)=2^(n-1)-1

Sn-S(n-1)=an

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

=2*2^(n-1)-2^(n-1)

=2^(n-1)

an^2=[2^(n-1)]^2

=2^(2n-2)

∴

a1^2=1

a2^2=2^2=4

a3^3=2^4=16

a4^2=2^6=64

a5^2=2^8=256

.

an^2=2^(2n-2)

希望对你能有所帮助.