A series whose n^{th} term is left( {frac{n}{ | vedclass

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Question

(a) On putting $n = 1,2,3,.....$

First term of the series $a = \frac{1}{x} + y$,

Second term =$\frac{2}{x} + y$

$d = \left( {\frac{2}{x} + y}

Vedclass · Accepted Answer

$\left\{ {\frac{{r(r + 1)}}{{2x}}} \right\} + ry$

Vedclass · Answer

(a) On putting $n = 1,2,3,.....$

First term of the series $a = \frac{1}{x} + y$,

Second term =$\frac{2}{x} + y$

$d = \left( {\frac{2}{x} + y} \right) - \left( {\frac{1}{x} + y} \right) = \frac{1}{x}$

Sum of $r$ terms of the series

$ = \frac{r}{2}\left[ {2\left( {\frac{1}{x} + y} \right) + (r - 1)\frac{1}{x}} \right]$

$ = \frac{r}{2}\left[ {\frac{2}{x} + 2y + \frac{r}{x} - \frac{1}{x}} \right]$

$ = \frac{{{r^2} - r + 2r}}{{2x}} + ry$

$ = \left\{ {\frac{{r{\mkern 1mu} (r + 1)}}{{2x}}} \right\} + ry$.

A series whose $n^{th}$ term is $\left( {\frac{n}{x}} \right) + y,$ the sum of $r$ terms will be

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