高一数学数列求和方面问题,急!数列{an}中,an=((-1)^n+4*n)/(2^n),求前n项和Sn要过程!
问题描述:
高一数学数列求和方面问题,急!
数列{an}中,an=((-1)^n+4*n)/(2^n),求前n项和Sn
要过程!
答
an=(-1/2)^n+4n/2^n
令Tn=4/2+4×2/2²+4×3/2³+。。。4×n/2^n①
则(1/2)Tn=4/2²+4×2/2³+。。。4×n/2^(n+1)②
①-②得(1/2)Tn=4(1/2+1/2²+1/2³+...1/2^n)-4n/2^(n+1)
TN=8(1-1/2^n)-4n/2^n
所以sn=((-1/2)^(n+1)+1/2)/(-1-1/2)+8(1-1/2^n)-4n/2^n =(1/3)(-1/2)^n+23/3-1/2^(n-3)-4n/2^n
答
∵a[n]=[(-1)^n+4n]/2^n=(-1/2)^n+4n/2^n∴我们先考察前一项:(-1/2)^n的前n项和T[n]即:T[n]=(-1/2)^1+(-1/2)^2+...+(-1/2)^n=(-1/2)[1-(-1/2)^n]/[1-(-1/2)]=(-1/3)[1-(-1/2)^n]我们再考察后一项:4n/2^n中n/2^n的...