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The sum of the series $\sum\limits_{r = 0}^n {{{( - 1)}^r}\,{\,^n}{C_r}\left( {\frac{1}{{{2^r}}} + \frac{{{3^r}}}{{{2^{2r}}}} + \frac{{{7^r}}}{{{2^{3r}}}} + \frac{{{{15}^r}}}{{{2^{4r}}}} + .....m\,{\rm{terms}}} \right)} $ is
$\frac{{{2^{mn}} - 1}}{{{2^{mn}}({2^n} - 1)}}$
$\frac{{{2^{mn}} - 1}}{{{2^n} - 1}}$
$\frac{{{2^{mn}} + 1}}{{{2^n} + 1}}$
None of these
Solution
(a) $\sum\limits_{r = 0}^n {{{( – 1)}^r}{\,^n}{C_r}\left( {\frac{1}{{{2^r}}} + \frac{{{3^r}}}{{{2^{2r}}}} + \frac{{{7^r}}}{{{2^{3r}}}} + ….\,m inksa rd} \right)} $
$ = \sum\limits_{r = 0}^n {{{( – 1)}^r}{\,^n}{C_r}} .\frac{1}{{{2^r}}} + \sum\limits_{r = 0}^n {{{( – 1)}^r}} {.^n}{C_r}\frac{{{3^r}}}{{{2^{2r}}}}$+ $\sum\limits_{r = 0}^n {{{( – 1)}^r}{\,^n}{C_r}} \frac{{{7^r}}}{{{2^{3r}}}} + …..$
$ = {\left( {1 – \frac{1}{2}} \right)^n} + {\left( {1 – \frac{3}{4}} \right)^n} + {\left( {1 – \frac{7}{8}} \right)^n} + ….$ upto $m$ terms
$ = \frac{1}{{{2^n}}} + \frac{1}{{{4^n}}} + \frac{1}{{{8^n}}} + ….$ upto $m$ terms
$ = \frac{{\frac{1}{{{2^n}}}\left( {1 – \frac{1}{{{2^{nm}}}}} \right)}}{{\left( {1 – \frac{1}{{{2^n}}}} \right)}} = \frac{{{2^{mn}} – 1}}{{{2^{mn}}({2^n} – 1)}}$
