sum a series - algebra

**thereddevils** · February 8th 2009, 01:05 AM

Find the sum of the series $2^2+4^2+6^2+...+50^2$

Using the above result , find the value of $1^2+3^2+5^2+...+49^2$ .

My working

$\sum^{25}_{r=1}4r^2=22100$

The question says use the above result and this is what i am confused with . How to use the result above to determine the sum of the following series . Is this the method of differences ? If yes , how do i apply ?

**Jhevon** · February 8th 2009, 01:33 AM

Originally Posted by thereddevils

Find the sum of the series $2^2+4^2+6^2+...+50^2$

Using the above result , find the value of $1^2+3^2+5^2+...+49^2$ .

My working

$\sum^{25}_{r=1}4r^2=22100$

nice

The question says use the above result and this is what i am confused with . How to use the result above to determine the sum of the following series . Is this the method of differences ? If yes , how do i apply ?

you know the formula for the sum of the first n squares. the first part is about the sum of the first even squares (that is, the squares of even integers), the second is asking for the sum of the first n odd squares. do you see what to do now? yes, finding a difference is involved

**thereddevils** · February 8th 2009, 02:16 AM

$u_r=4r^2$ but i don see that $u_r=f(r)-f(r-1)$ , so how can i apply the method of differences?

Thanks again ,Ｊｈｅｖｏｎ　．　

**mr fantastic** · February 8th 2009, 02:32 AM

Originally Posted by thereddevils

$u_r=4r^2$ but i don see that $u_r=f(r)-f(r-1)$ , so how can i apply the method of differences?

Thanks again ,Ｊｈｅｖｏｎ　．　

$S = \sum_{r = 1}^{50} r^2 = 1^2 + 2^2 + 3^2 + \, .... \, + \, 49^2 + 50^2$ can be calculated from the standard formula.

But $S = \sum_{r = 1}^{50} r^2 = (1^2 + 3^2 + \, .... \, + 49^2) + (2^2 + 4^2 + \, .... \, + 50^2)$ .

Therefore $(1^2 + 3^2 + \, .... \, + 49^2) = \sum_{r = 1}^{50} r^2 - (2^2 + 4^2 + \, .... \, + 50^2) = S - 22100 = \, ....$

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