Question

求和:(1/2²-1)+(1/3²-1)+(1/4²-1)+....+(1/n²-1)

Answer 1

∵1/﹙n²-1﹚=2/﹙n-1﹚-2/﹙n+1﹚
∴(1/2²-1)+(1/3²-1)+(1/4²-1)+....+(1/n²-1)
=(2/1×3)+(2/2×4)+(2/3×5)+....+﹛2/﹙n-1)﹙n+1﹚﹜
=2/1-2/3+2/2-2/4+2/3-2/5+.....+2/﹙n-1)-2/﹙n+1﹚
=2+1－2/n－2/﹙n+1﹚
=﹙3n²+n-1﹚／[n﹙n+1﹚]

Answer 2

(1/2²-1)+(1/3²-1)+(1/4²-1)+....+(1/n²-1)
=[1-1/3+1/2-1/4+1/3-1/6+……+1/(n-1)-1/(n+1)]÷2
=[1+1/2-1/n-1/(n+1)]÷2
=(3n²+n-1)/(2n²+2n)