Author Topic: Sum of a series  (Read 2179 times)

Offline Ghost Of Highbury

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Sum of a series
« on: September 03, 2011, 06:47:24 pm »
I need a formula to find the sum of this series.

2,6,12,20....

the nth term is n(n+1)

divine intervention!

Amelia

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Re: Sum of a series
« Reply #1 on: September 04, 2011, 04:38:03 pm »
Answers only work once. In exams it is methods that count.

To find the sum of the series 2,6,12,20... given nth term is n(n+1)
=======================================…
Method 1: Experimentation
Compare partial sums with individual terms
2, 8, 20, 40, 70, 112
2, 6, 12, 20, 30, 42

Since denominator is often 3 express all these in
terms of thirds and observe how ratio depends on n

3/3, 4/3, 5/3, 6/3, 7/3, 8/3

3 is 1 +2,
4 is 2 + 2
5 is 3 + 2 and so on.

Sum to n terms, S = [(n + 2)/3] * nth term
Sum to n terms, S = n(n+1)(n + 2)/3
=========================

Although that was easy to understand, it relied on the (unlikely?) chance of finding a simple relation between partial sums and nth term. A more reliable approach uses the SIGMA notation.

Method 2: The ith term is i^2 + i so what we want is the sum S, where
S = (i = 1 to n)SIGMA [i^2 + i]. The 1 to n part is understood in the next few SIGMAS

SIGMA[(i + 1)^3 - i^3] = SIGMA(3i^2 + 3i) + SIGMA (1) = 3S + n ...(a)
The justification of introducing that quantity is that it "telescopes"

SIGMA[(i + 1)^3 - i^3] = (2^3 - 1^3) + (3^3 - 2^3) + (4^3 - 3^3) + ... + [(n + 1)^3 - n^3]
so that all terms except the first and last cancel out

SIGMA[(i + 1)^3 - i^3] = (n + 1)^3 - 1 ........................(b)

Comparing a) and b)
3S = n^3 + 3n^2 + 3n + 1 - 1 - n = n^3 + 3n^2 + 2n which rearranges as
Sum to n terms, S = n(n+1)(n + 2)/3
=======================

Regards - Ian (YA)

Offline ~ Miss Relina ~

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Re: Sum of a series
« Reply #2 on: September 04, 2011, 05:35:18 pm »
Answers only work once. In exams it is methods that count.

To find the sum of the series 2,6,12,20... given nth term is n(n+1)
=======================================…
Method 1: Experimentation
Compare partial sums with individual terms
2, 8, 20, 40, 70, 112
2, 6, 12, 20, 30, 42

Since denominator is often 3 express all these in
terms of thirds and observe how ratio depends on n

3/3, 4/3, 5/3, 6/3, 7/3, 8/3

3 is 1 +2,
4 is 2 + 2
5 is 3 + 2 and so on.

Sum to n terms, S = [(n + 2)/3] * nth term
Sum to n terms, S = n(n+1)(n + 2)/3
=========================

Although that was easy to understand, it relied on the (unlikely?) chance of finding a simple relation between partial sums and nth term. A more reliable approach uses the SIGMA notation.

Method 2: The ith term is i^2 + i so what we want is the sum S, where
S = (i = 1 to n)SIGMA [i^2 + i]. The 1 to n part is understood in the next few SIGMAS

SIGMA[(i + 1)^3 - i^3] = SIGMA(3i^2 + 3i) + SIGMA (1) = 3S + n ...(a)
The justification of introducing that quantity is that it "telescopes"

SIGMA[(i + 1)^3 - i^3] = (2^3 - 1^3) + (3^3 - 2^3) + (4^3 - 3^3) + ... + [(n + 1)^3 - n^3]
so that all terms except the first and last cancel out

SIGMA[(i + 1)^3 - i^3] = (n + 1)^3 - 1 ........................(b)

Comparing a) and b)
3S = n^3 + 3n^2 + 3n + 1 - 1 - n = n^3 + 3n^2 + 2n which rearranges as
Sum to n terms, S = n(n+1)(n + 2)/3
=======================

Regards - Ian (YA)

excuse me is that explanation should be known for C1 or not ???
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Offline Tohru Kyo Sohma

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Re: Sum of a series
« Reply #3 on: September 04, 2011, 06:25:15 pm »
Answers only work once. In exams it is methods that count.

To find the sum of the series 2,6,12,20... given nth term is n(n+1)
=======================================…
Method 1: Experimentation
Compare partial sums with individual terms
2, 8, 20, 40, 70, 112
2, 6, 12, 20, 30, 42

Since denominator is often 3 express all these in
terms of thirds and observe how ratio depends on n

3/3, 4/3, 5/3, 6/3, 7/3, 8/3

3 is 1 +2,
4 is 2 + 2
5 is 3 + 2 and so on.

Sum to n terms, S = [(n + 2)/3] * nth term
Sum to n terms, S = n(n+1)(n + 2)/3
=========================

Although that was easy to understand, it relied on the (unlikely?) chance of finding a simple relation between partial sums and nth term. A more reliable approach uses the SIGMA notation.

Method 2: The ith term is i^2 + i so what we want is the sum S, where
S = (i = 1 to n)SIGMA [i^2 + i]. The 1 to n part is understood in the next few SIGMAS

SIGMA[(i + 1)^3 - i^3] = SIGMA(3i^2 + 3i) + SIGMA (1) = 3S + n ...(a)
The justification of introducing that quantity is that it "telescopes"

SIGMA[(i + 1)^3 - i^3] = (2^3 - 1^3) + (3^3 - 2^3) + (4^3 - 3^3) + ... + [(n + 1)^3 - n^3]
so that all terms except the first and last cancel out

SIGMA[(i + 1)^3 - i^3] = (n + 1)^3 - 1 ........................(b)

Comparing a) and b)
3S = n^3 + 3n^2 + 3n + 1 - 1 - n = n^3 + 3n^2 + 2n which rearranges as
Sum to n terms, S = n(n+1)(n + 2)/3
=======================

Regards - Ian (YA)
errr
i was never good with this topic
plus i didn't understand a word
but hats off for a fine answer which is correct(i can't understand...but its looks very right!)
:P

Amelia

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Re: Sum of a series
« Reply #4 on: September 04, 2011, 08:01:02 pm »
Math Geeks are not posting  ::) This is not my work, I got it from Yahoo Answers.

I apologize for all the confusion.  ::)

Offline Tohru Kyo Sohma

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Re: Sum of a series
« Reply #5 on: September 04, 2011, 08:36:17 pm »
Math Geeks are not posting  ::) This is not my work, I got it from Yahoo Answers.

I apologize for all the confusion.  ::)
oh
okie!
confusion cleared!

Offline The Golden Girl =D

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Re: Sum of a series
« Reply #6 on: September 04, 2011, 09:16:56 pm »
Math Geeks are not posting  ::) This is not my work, I got it from Yahoo Answers.

I apologize for all the confusion.  ::)

Haven't touched a Math book for quite sometime :-\  Sorry I can't be of help :(
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