Question

数列2/2,4/2^2,6/2^3,……,2n/2^n,……的前n项的和sn=

Answer 1

Sn =2/2 + 4/2² + 6/2³ + …… + 2(n-1)/2^(n-1) + 2n/2^n …………①

2Sn=2×(2/2 + 4/2² + 6/2³ + …… + 2n/2^n)

=2 + 4/2 + 6/2² + 8/2³ + …… + 2n/2^(n-1) …………②

②减①得

Sn=2 + 2/2 + 2/2² + 2/2³ + …… +2/2^(n-1) - 2n/2^n

=2[1-(1/2)^n]/[1-(1/2)] - 2n/2^n

=4 - 4/2^n - 2n/2^n

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

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

这叫错位相减法