Question

高二数列问题

An=n乘以2的n次方前n项求和

Answer 1

an=n*2^n

a1=1*2^1
a2=2*2^2
a3=3*2^3
....
an=n*2^n

Sn=1*2^1+2*2^2+3*2^3+....+n*2^n
2Sn=1*2^2+2*2^3+3*2^4+....+(n-1)*2^n+n*2^(n+1)
Sn-2Sn=2^1+2^2+3*2^3+....+2^n-n*2^(n+1)
=2*(1-2^n)/(1-2)-n*2^(n+1)
=2^(n+1)-2-n*2^(n+1)
=(1-n)*2^(n+1)-2
=-(n-1)*2^(n+1)-2
Sn=(n-1)*2^(n+1)+2