Question

数列求和:Sn=1/2^2-1+1/3^2-1+1/4^2-1+.....+1/n^2-1 麻烦各位帮忙解一下...急用

Answer 1

Sn=1/2^2 - 1 + 1/3^2 - 1 + 1/4^2 - 1 +..... + 1/n^2 - 1
=1/(2 - 1)(2 + 1) + 1/(3 - 1)(3 + 1) + ...... + 1/(n - 1)(n + 1)
=1/(1 * 3) + 1/(2 * 4) + ...... + 1/(n - 1)(n + 1)
=1/2[1/1 - 1/3 + 1/2 - 1/4 + ...... + 1/(n - 1)(n + 1)]
=1/2[1 + 1/2 - 1/n - 1/(n + 1)]
=3/4 - 1/(2n) - 1/(2n + 2)

Answer 2

因为an=1/2(1/(n-1)-1/(n+1))
所以sn=a1+a2+.....+an
=1/2(1-1/3+1/2-1/4+1/3-1/5+.....+1/(n-1)-1/(n+1))
=1/2(1+1/2-1/n-1/(n+1))
=1/2(3/2-1/n-1/(n+1))
=(3*n*n-n-2)/(4n(n+1))