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The value of ${\sum\limits_{r = 1}^{19} {\frac{{{}^{20}{C_{r + 1}}\left( { - 1} \right)}}{{{2^{2r + 1}}}}} ^r}$ is
$2\left( {{{\left( {\frac{3}{4}} \right)}^{20}} + 4} \right)$
$-2\left( {{{\left( {\frac{3}{4}} \right)}^{20}} + 4} \right)$
$2\left( {{{\left( {\frac{3}{4}} \right)}^{20}} - 4} \right)$
$-2\left( {{{\left( {\frac{3}{4}} \right)}^{20}} - 4} \right)$
Solution
$\frac{1}{2}\sum\limits_{r = 1}^{19} {^{20}{C_{r + 1}}} {\left( {\frac{{ – 1}}{4}} \right)^r} = \frac{1}{2}{\sum\limits_{r = 2}^{20} {^{20}{C_r}\left( { – \frac{1}{4}} \right)} ^{r – 1}}$
$ = – 2{\sum\limits_{r = 2}^{20} {^{20}{C_r}\left( { – \frac{1}{4}} \right)} ^r}$
$=-2\left\{\left(1-\frac{1}{4}\right)^{20}-\left(^{20} \mathrm{C}_{0}\left(-\frac{1}{4}\right)^{0}+^{20} \mathrm{C}_{1}\left(-\frac{1}{4}\right)^{1}\right)\right\}$
$=-2\left\{\left(\frac{3}{4}\right)^{20}+4\right\}$
