Question

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！

Answer 1

解：∵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的前n项和R[n]，其中系数4最后再处理
∵R[n]=1/2^1+2/2^2+3/2^3+...+n/2^n
∴R[n]/2=1/2^2+2/2^3+3/2^4+...+(n-1)/2^n+n/2^(n+1)
∴上面两式相减：
R[n]/2=1/2^1+1/2^2+1/2^3+...+1/2^n-n/2^(n+1)
=1/2[(1-1/2^n)/(1-1/2)]-n/2^(n+1)
=1-1/2^n-n/2^(n+1)
=1-(n/2+1)/2^n
∴R[n]=2-(n+2)/2^n

综上所述：
S[n]=T[n]+4R[n]
=(-1/3)[1-(-1/2)^n]+8-4(n+2)/2^n
=23/3+(1/3)(-1/2)^n-4(n+2)/2^n
=23/3+[(1/3)(-1)^n-4(n+2)]/2^n

Answer 2

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