IIT JEE , AIEEE Forum - Mathematics , Algebra

tuhinr_007

Find the sum of n terms of the series 1/3+3/(3.7)+5/(3.7.11)+7/(3.7.11.15)+...................

amulye

wats d ans given plz tell me

tuhinr_007

The answer given is--- 1/2-1/(2.3.7.11........(4n-1))

(I think that there is some problem with the answer given in the buk.

watz ur answer???

tuhinr_007

pls answer

tuhinr_007

nobody!!!!!!!!!!

sachinguptaiit

$T_r=\frac{2r-1}{\left(\displaystyle\prod_{r=1}^n(4r-1)\right)}$

$T_r=\left(\frac{1}{2}\right)\left(\frac{(4r-2)}{\left(\displaystyle\prod_{r=1}^n(4r-1)\right)}\right)$

$T_r=\left(\frac{1}{2}\right)\left(\frac{(4r-1)-(1)}{\left(\displaystyle\prod_{r=1}^n(4r-1)\right)}\right)$

$T_r=\left(\frac{1}{2}\right)\left(\frac{1}{\left(\displaystyle\prod_{r=1}^n(4r-5)\right)}-\frac{1}{\left(\displaystyle\prod_{r=1}^n(4r-1)\right)}\right)$

$\displaystyle\sum_{i=1}^{n}T_r=\displaystyle\sum_{i=1}^{n}\left(\frac{1}{2}\right)\left(\frac{1}{\left(\displaystyle\prod_{r=1}^n(4r-5)\right)}-\frac{1}{\left(\displaystyle\prod_{r=1}^n(4r-1)\right)}\right)$

$\text{Since these are continuous terms,we put n=1}\\\\\text{in earlier term and put n=n in other term }$

$\displaystyle\sum_{i=1}^{n}T_r=\left(\frac{1}{2}\right)\left(\frac{1}{1}-\frac{1}{\left(\displaystyle\prod_{r=1}^n(4r-1)\right)}\right)$

$S_n=\left(\frac{1}{2}\right)\left(\frac{1}{1}-\frac{1}{\left(\displaystyle\prod_{r=1}^n(4r-1)\right)}\right)$

$S_n=\frac{1}{2}-\frac{1}{2.3.7.11..(4n-1)}$

$\text{Hence,Proved}$

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