श्रेणी sumlimits_{r = 0}^n {{{( - 1)}^r},{,^n | vedclass

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Question

(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

Vedclass · Accepted Answer

$\frac{{{2^{mn}} - 1}}{{{2^{mn}}({2^n} - 1)}}$

Vedclass · Answer

(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} + ....$ $m$ पदों तक

$ = \frac{1}{{{2^n}}} + \frac{1}{{{4^n}}} + \frac{1}{{{8^n}}} + ....$ $m$ पदों तक

$ = \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)}}$

श्रेणी $\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\,inksa rd } \right)} $ का योगफल है

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