The value of sumlimits_{r = 1}^{19} {frac{{{} | vedclass

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(B) Let $S = \sum\limits_{r = 1}^{19} \frac{{}^{20}C_{r+1}(-1)^r}{2^{2r+1}}$.We can rewrite the term as $\frac{1}{2} \sum\limits_{r=1}^{19} {}^{20}C_{

Vedclass · Accepted Answer

$-2\left( {{{\left( {\frac{3}{4}} \right)}^{20}} + 4} \right)$

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(B) Let $S = \sum\limits_{r = 1}^{19} \frac{{}^{20}C_{r+1}(-1)^r}{2^{2r+1}}$.We can rewrite the term as $\frac{1}{2} \sum\limits_{r=1}^{19} {}^{20}C_{r+1} \left(-\frac{1}{4}\right)^r$.Let $k = r+1$,then as $r$ goes from $1$ to $19$,$k$ goes from $2$ to $20$.$S = \frac{1}{2} \sum\limits_{k=2}^{20} {}^{20}C_k \left(-\frac{1}{4}\right)^{k-1} = \frac{1}{2} \sum\limits_{k=2}^{20} {}^{20}C_k (-4) \left(-\frac{1}{4}\right)^k = -2 \sum\limits_{k=2}^{20} {}^{20}C_k \left(-\frac{1}{4}\right)^k$.Using the binomial expansion $(1+x)^n = \sum\limits_{k=0}^n {}^{n}C_k x^k$,we have $\sum\limits_{k=2}^{20} {}^{20}C_k \left(-\frac{1}{4}\right)^k = (1 - \frac{1}{4})^{20} - {}^{20}C_0 - {}^{20}C_1(-\frac{1}{4})$.$= (\frac{3}{4})^{20} - 1 - 20(-\frac{1}{4}) = (\frac{3}{4})^{20} - 1 + 5 = (\frac{3}{4})^{20} + 4$.Therefore,$S = -2 \left( (\frac{3}{4})^{20} + 4 \right)$.

The value of $\sum\limits_{r = 1}^{19} {\frac{{{}^{20}{C_{r + 1}}{(-1)}^r}}{{{2^{2r + 1}}}}}$ is

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