The first term of an arithmetic series is 240 and common diff is -8. Where the nth term is Un (u sub n)
a) Write an expression for Un
b) Given that Uk=0, find the value of (Sigma notation) with K at the top and n=1 on bottom. with Un to the side....
For a) i wrote 240 + (n-1)-8?
Arithmetic series using sigma notation help?
2013-05-05 6:11 pm
回答 (2)
2013-05-05 6:28 pm
Your answer for (a) is correct if I assume you meant (n-1)(-8) so that you are multiplying the two., You can then simplify it further.
Un = 240 + (n-1)(-8) = 240 -8n + 8 = 248 - 8n.
You want to find k such that Uk = 0. Just plug into your general formula
Uk = 248 -8k = 0
8k = 248
k = 248/8 = 31
Now you want to find the sum
31
∑ 248 - 8n =
n=1
Now you can break this into two sums
31
∑248 - 8∑n =
n=1
The first sum simply gives you (31)(248).
I presume the ∑ n is well know to you. The result is
(31)(248) - 8N(N+1)/2 = (31)(248) - (8)(31)(32)/2 = (31)[248 - (8)(16)] = 31(120) = 3720
Un = 240 + (n-1)(-8) = 240 -8n + 8 = 248 - 8n.
You want to find k such that Uk = 0. Just plug into your general formula
Uk = 248 -8k = 0
8k = 248
k = 248/8 = 31
Now you want to find the sum
31
∑ 248 - 8n =
n=1
Now you can break this into two sums
31
∑248 - 8∑n =
n=1
The first sum simply gives you (31)(248).
I presume the ∑ n is well know to you. The result is
(31)(248) - 8N(N+1)/2 = (31)(248) - (8)(31)(32)/2 = (31)[248 - (8)(16)] = 31(120) = 3720
2013-05-05 6:19 pm
(a) U1 = 240, u2 = 240 - 8; U3 = (240 - 8) - 8 = 240 - 2*8, U4 = 240 - 3*8 so Un = 240 - 8(n-1) = 240 - 8n + 8 so Un = 248 - 8n. (b) If Uk = 0, put n = k and 248 - 8k = 0 so 8k = 240, k = 30. If I understand you correctly, you want Sigma(n=1 at the bottom, k at the top)[248 - 8n] = 0
參考: Retired Maths Teacher
收錄日期: 2021-05-01 14:48:40
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20130505101142AAk33X0