Question

数列 1*1/2,2*1/4,3*1/8,4*1/16, ...的前n项和为？求解题步骤

Answer 1

Sn=1*1/2+2*1/2²+3*1/2的三次方+...+n*1/2的n次方
1/2Sn=1*1/2²+2*1/2的三次方+...+n*1/2的（n+1）次方
相减得1/2Sn=(1/2+1/2²+1/2的三次方+...+1/2的n次方)+1/2的（n+1)次方
括号中为等比数列，自己应该会算

Answer 2

Sn=1*1/2+2*1/4+…+n*1/2^n
1/2*Sn=1*1/4+2*1/8+…+n*1/2^(n+1)
相减
1/2*Sn=1/2+1/4+……+1/2^n-n*1/2^(n+1)
=1/2*[1-(1/2)^n]/(1-1/2)-n*1/2^(n+1)
=1-1/2^n-n*1/2^(n+1)
Sn=2-1/2^(n-1)-n*1/2^n

Answer 3

s= 1*1/2+2*1/4+3*1/8+ ...+(n-1)/2^(n-1)+ n/2^n
2s = 1*1+2*1/2+3*1/4+4*1/8+ ...+ n/2^(n-1)
s = 2s-s = 1 +[1/2+1/4+1/8+ ...+ 1/2^(n-1)]- n/2^n
=1+[1/2 - 1/2^n]/[1-1/2]- n/2^n
=2-1/2^(n-1) -n/2^n

Answer 4

设为S
0.5S=S-0.5S=1*1/2+(2-1)*1/4+(3-2)*1/8...

Answer 5

通项an=n*(1/2)^n
Sn=1*1/2+2*1/2^2+...+n*1/2^n
2Sn=1*1+2*1/2+...+n*1/2^(n-1)
2Sn-Sn=1+1/2+...+1/2^(n-1)-n*1/2^n
所以,Sn=1*(1-(1/2)^n)/(1-1/2)-n*1/2^n=2(1-1/2^n)-n*1/2^n=2-(n+2)*1/2^n