Question

已知等差数列{1/(2n+1)(2n+3)},求此数列的前n项和

Answer 1

解1/（2n+1）*（2n+3）=1/2(1/（2n+1）-1/（2n+3）)
1/3*5=1/2(1/3-1/5)
1/5*7=1/2(1/5-1/7)
1/7*9=1/2(1/7-1/9)
.......................
1/（2n+1）*（2n+3）=1/2(1/（2n+1）-1/（2n+3）)
上述各式相加得
1/2(1/3-1/5)+1/2(1/5-1/7)+1/2(1/7-1/9)+.........++1/2(1/（2n+1）-1/（2n+3）)
=1/2(1/3-1/5+1/5-1/7+1/7-1/9+.........+(1/（2n+1）-1/（2n+3）)
=1/2(1/3-1/2n+3)
=1/6-1/2（2n+3）

追问

麻烦问一下，第一步的那个二分之一是如何带入的

追答

解1/2*(1/（2n+1）-1/（2n+3）)
=1/2*[（2n+3）/（2n+1）*（2n+3）-（2n+1）/（2n+1）（2n+3）]
=1/2*{[（2n+3）-（2n+1）]/（2n+1）（2n+3）}
=1/2*{2/（2n+1）（2n+3）}
=1/（2n+1）*（2n+3）
即1/（2n+1）*（2n+3）=1/2(1/（2n+1）-1/（2n+3）)

Answer 2

解答：
裂项求和
2/(2n+1)(2n+3)
=[(2n+3)-(2n+1)]/(2n+1)(2n+3)
=1/(2n+1)-1/(2n+3)
∴ 1/(2n+1)(2n+3)=(1/2)*[1/(2n+1)-1/(2n+3)]
∴ 设前n项和为Sn
∴ Sn=(1/2)(1/3-1/5)+(1/2)(1/5-1/7)+......+(1/2)*[1/(2n+1)-1/(2n+3)]
=(1/2)[1/3-1/(2n+3)]
=(1/2)(2n+3-3)/ [3*(2n+3)]
=n/(6n+9)

Answer 3

前n项和
=1/3×5＋1/5×7＋1/7×9＋.....＋1/(2n+1)(2n+3)
=1/2 [1/3-1/5＋1/5-1/7＋1/7-1/9＋。。。＋1/(2n+1)-1/(2n+3)]
=1/2[1/3-1/(2n+3)]
=1/2[2n/(2n+3)]
=n/(2n+3)