Question

已知数列a[n]=2n-1,b[n]=2^n,令c1=1,c[2k]=a[2k-1],c[2k+1]=a[2k]+kb[

Answer 1

由已知得,c[2k]=4k-3,c[2k+1]=4k-1+k*2^k,把T[2n+1]分成两部分求和,奇数项和偶数项分别求.

偶数项的和为：4(n+1)n/2-3n；奇数项的和为：4(n+1)n/2 -n-1+2^1+2*2^2+3*2^3+……+n*2^n,

令X=2^1 +2*2^2 +3*2^3 +……+n*2^n

则2X= 2^2 +2*2^3 +……+(n-1)*2^n +n*2^(n+1)

两式相减得到X= -2^1 -2^2 -2^3 -…… -2^n +n*2^(n+1)= 2(1-2^n) +n*2^(n+1);

所以奇数项的和为：4(n+1)n/2 -n-1+2(1-2^n) +n*2^(n+1)

则T[2n+1]=4(n+1)n/2-3n +4(n+1)n/2 -n-1+2(1-2^n) +n*2^(n+1)

=4n^2 +(n-1)2^(n+1) +1