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# Find the sum of the first 20 terms of the series: 1^2 + 2^2

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Manager
Joined: 25 Jul 2006
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Find the sum of the first 20 terms of the series: 1^2 + 2^2 [#permalink]  04 Jul 2007, 16:25
Find the sum of the first 20 terms of the series:
1^2 + 2^2 + 3^2 + ... + 20^2

Does anyone know of a formula? Thanks.
Manager
Joined: 11 Mar 2007
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I'm pretty sure it's:

(n*(n+1)*(2n+1))/6

So if we had use 4, it would be (4*5*9)/6 = 180/6 = 30

1 + 4 + 9 + 16 = 30

20 would be:

(20*21*41)/6 = 2870

You may want to check that...
Manager
Joined: 25 Jul 2006
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mavery wrote:
I'm pretty sure it's:

(n*(n+1)*(2n+1))/6

So if we had use 4, it would be (4*5*9)/6 = 180/6 = 30

1 + 4 + 9 + 16 = 30

20 would be:

(20*21*41)/6 = 2870

You may want to check that...

wonderful - thanks!
seems to work for some no.s i tried. can you plz point me to some definitive reference on how this formula is arrived at - couldn't find it online.

also, is there a variation of this formula that is based on first term, difference, etc. (like arithmetic series Sum formula).

For example, is there a way to solve?

1. 1^2 + 3^2 + 5^2 + ...19^2

2. 4^2 + 7^2 + 10^2 + ...25^2
Manager
Joined: 11 Mar 2007
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I found it in some notes I found online...I've attached the file...

As for the other question...

I've ran across this problem that you may be able to tweak to meet your needs...

1. The sum of the even numbers between 1 and n is 79*80, where n is an odd number, then n=?

Sol: First term a=2, common difference d=2 since even number

therefore sum to first n numbers of Arithmetic progression would be

n/2(2a+(n-1)d)

= n/2(2*2+(n-1)*2)=n(n+1) and this is equal to 79*80

therefore n=79 which is odd...
Attachments

General Math 1.doc [120 KiB]

Senior Manager
Joined: 04 Jun 2007
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mavery wrote:
I found it in some notes I found online...I've attached the file...

As for the other question...

I've ran across this problem that you may be able to tweak to meet your needs...

1. The sum of the even numbers between 1 and n is 79*80, where n is an odd number, then n=?

Sol: First term a=2, common difference d=2 since even number

therefore sum to first n numbers of Arithmetic progression would be

n/2(2a+(n-1)d)

= n/2(2*2+(n-1)*2)=n(n+1) and this is equal to 79*80

therefore n=79 which is odd...

n should really be 159 !
79 is just the number of terms in the sequence !!

And thanks for the doc.
Manager
Joined: 25 Jul 2006
Posts: 97
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1. thx for the doc - it has a formula for the sum of cubes series, which also came up for me - useful!

2. the example given is a simple linear AP (not involving squares or cubes) - not tweakable for sq or cube series (yes, agree - the ans is 159)
regardless, thanks!
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