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A series whose $n^{th}$ term is $\left( {\frac{n}{x}} \right) + y,$ the sum of $r$ terms will be
$\left\{ {\frac{{r(r + 1)}}{{2x}}} \right\} + ry$
$\left\{ {\frac{{r(r - 1)}}{{2x}}} \right\}$
$\left\{ {\frac{{r(r - 1)}}{{2x}}} \right\} - ry$
$\left\{ {\frac{{r(r + 1)}}{{2y}}} \right\} - rx$
Solution
(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$.
