若an=-2n+21数列{bn}满足bn=ancos(nπ)+2^n求数列{bn}的前n项和

Question

若an=-2n+21数列{bn}满足bn=ancos(nπ)+2^n求数列{bn}的前n项和

Answer 1

cos(nπ)=(-1)ⁿ
bn=ancos(nπ)+2ⁿ
=(-2n+21)(-1)ⁿ+2ⁿ
=21×(-1)ⁿ-2n×(-1)ⁿ+2ⁿ
Tn=b1+b2+...+bn
=21×[(-1)+(-1)²+...+(-1)ⁿ]-2[1×(-1)+2×(-1)²+...+n×(-1)ⁿ]+(2+2²+...+2ⁿ)
2+2²+...+2ⁿ=2(2ⁿ-1)/(2-1)=2^(n+1) -2
n为奇数时,
(-1)+(-1)²+...+(-1)ⁿ=-1
1×(-1)+2×(-1)²+...+n×(-1)ⁿ=-1+2-3+4-...-(n-2)+(n-1)-n=(n-1)/2 -n=-(n+1)/2
Tn=-21+(n+1)+2 +2^(n+1) -2=2^(n+1) +n+22
n为偶数时,
(-1)+(-1)²+...+(-1)ⁿ=0
1×(-1)+2×(-1)²+...+n×(-1)ⁿ=-1+2-3+4-...-(n-1)+n=n/2
Tn=0-n+2^(n+1)-2=2^(n+1) -n-2
写成统一的形式：
Tn=2^(n+1) -(n+10)(-1)ⁿ+12