Question

千卉

請求大神教教我

2. 化簡
1 1
+
3×77x11
+
1
11×15
+‥+
1
(4n-1)(4n+3)
。

Answer 1

qn

(對於 1≤k≤n)
1/[(4k-1)(4k+3)]
=¼ × 4/[(4k-1)(4k+3)]
=¼ × [(4k+3)-(4k-1)]/[(4k-1)(4k+3)]
=¼ × { (4k+3)/[(4k-1)(4k+3)] - (4k-1)/[(4k-1)(4k+3)] }
=¼ × { 1/(4k-1) - 1/(4k+3) }

所以
1/(3×7) + 1/(7×11) + 1/(11×15) + ⋯ + 1/[(4n-1)(4n+3)]
=¼[1/3 - 1/7]+¼[1/7 - 1/11]+¼[1/11 - 1/15]+⋯+¼[1/(4n-1) - 1/(4n+3)]
=¼×(1/3) - ¼×[1/(4n+3)]
=¼ × [(4n+3)-3]/[3(4n+3)]
= n/(12n+9)