Cn=n+【1/(2^n)】求数列前n项和Sn

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Cn=n+【1/(2^n)】求数列前n项和SnCn=n+【1/(2^n)】求数列前n项和SnCn=n+【1/(2^n)】求数列前n项和Sn∵c[n]=n+1/2^n∴S[n]=(1+1/2^1)+(2

Cn=n+【1/(2^n)】求数列前n项和Sn
Cn=n+【1/(2^n)】求数列前n项和Sn

Cn=n+【1/(2^n)】求数列前n项和Sn
∵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