Question

Cn=n+（1/（2^n））求数列前n项和Sn

Answer 1

∵c[n]=n+1/2^n

∴S[n]=(1+1/2^1)+(2+1/2^2)+(3+1/2^3)+...+(n+1/2^n)

=(1+2+3+...+n)+(1/2^1+1/2^2+1/2^3+...+1/2^n)

=n(n+1)/2+(1/2)[1-(1/2)^n]/(1-1/2)

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