Q) Find the sum of the series: a). 1/(2^1/2 + 1^1/2) + 1/(2^1/2 + 3^1/2) + .......+ 1/(120^1/2 +121^1/2) b). 1/(1 x 5) + 1/(5 x 9) + 1/(9 x 13) + .......+ 1/(221 x 225). Any help would be appreciated. Thanks
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Originally Posted by a69356 Q) Find the sum of the series: a). 1/(2^1/2 + 1^1/2) + 1/(2^1/2 + 3^1/2) + .......+ 1/(120^1/2 +121^1/2) b). 1/(1 x 5) + 1/(5 x 9) + 1/(9 x 13) + .......+ 1/(221 x 225). These are both telescoping series. For a), the n'th term is . For b), the n'th term is , which you can split into partial fractions.
Thanks Opalg, I found the sum of the first series it comes out to be 10. But for the second one I found that there are 56 terms in the series and the series is neither A.P nor G.P. How do we proceed further to find the sum of it. Thanks, Ashish
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