负一的n次方乘以(22-2n),这个数列的前n项和
答案:1 悬赏:70 手机版
解决时间 2021-01-31 17:18
- 提问者网友：佞臣
- 2021-01-31 01:43
负一的n次方乘以(22-2n),这个数列的前n项和
最佳答案
- 五星知识达人网友：有你哪都是故乡
- 2021-01-31 02:31
an=(-1)^n .(22-2n)
an = 2n-22 ; n is odd
= 22-2n ; n is even
Sn = a1+a2+...+an
case 1: if n is odd
Sn = (a1+a3+....+an) + (a2+a4+...+a(n-1))
a1+a3+...+an
=(2+6+10+...+2n) - 11(n+1)
= (1/2)(n+1)^2 -11(n+1)
a2+a4+...+a(n-1)
=11(n-1) - (2+4+...+2(n-1))
=11(n-1) - (1/2)n(n-1)
=>
Sn = (1/2)(n+1)^2 -11(n+1) +11(n-1) - (1/2)n(n-1)
= (1/2)(3n+1)-22
case 2: if n is even
Sn = (a1+a3+....+a(n-1)) + (a2+a4+...+an)
a1+a3+...+a(n-1)
=(2+6+10+...+2(n-1)) - 11n
= (1/2)n^2 -11n
a2+a4+...+an
=11n - (2+4+...+2n)
=11(n-1) - (1/2)n(n+1)
Sn =(1/2)n^2 -11n + 11(n-1) - (1/2)n(n+1)
=-(1/2)n-11
an = 2n-22 ; n is odd
= 22-2n ; n is even
Sn = a1+a2+...+an
case 1: if n is odd
Sn = (a1+a3+....+an) + (a2+a4+...+a(n-1))
a1+a3+...+an
=(2+6+10+...+2n) - 11(n+1)
= (1/2)(n+1)^2 -11(n+1)
a2+a4+...+a(n-1)
=11(n-1) - (2+4+...+2(n-1))
=11(n-1) - (1/2)n(n-1)
=>
Sn = (1/2)(n+1)^2 -11(n+1) +11(n-1) - (1/2)n(n-1)
= (1/2)(3n+1)-22
case 2: if n is even
Sn = (a1+a3+....+a(n-1)) + (a2+a4+...+an)
a1+a3+...+a(n-1)
=(2+6+10+...+2(n-1)) - 11n
= (1/2)n^2 -11n
a2+a4+...+an
=11n - (2+4+...+2n)
=11(n-1) - (1/2)n(n+1)
Sn =(1/2)n^2 -11n + 11(n-1) - (1/2)n(n+1)
=-(1/2)n-11
我要举报
如以上问答信息为低俗、色情、不良、暴力、侵权、涉及违法等信息，可以点下面链接进行举报！
大家都在看
推荐资讯