Question

求这数列前n项和bn＝n/2的2n－1次方,求bn前n项和,详细.

Answer 1

令S=b1+b2+b3+.+bn

=1/2^1+2/2^3+3/2^5+.n/2^(2n-1)

s/4=1/2^3+2/2^5+3/2^7+.n/2^(2n+1)

s-s/4=1/2^1+1/2^3+1/2^5+.+1/2^(2n-1)-n/2^(2n+1)

3s/4=1/2+1/8*[(1-1/4^(n-1)/(1-1/4)]-n/2^(2n+1)

s=2/3+2/9*[(1-1/4^(n-1)]-n/2^(2n+1)