Question

The sum of the series $2·{}^{20}{C}_0+5{}^{20}{C}_1+8{}^{20}{C}_2+...........+62{}^{20}{C}_{20}$ is equal to

• A. $16·2^{22}$
• B. $8·2^{22}$
• C. $8·2^{21}$
• D. $16·2^{21}$

Answer 1

The correct option is D

$16·2^{21}$

Explanation for the correct option

Given: $2·{}^{20}{C}_0+5{}^{20}{C}_1+8{}^{20}{C}_2+...........+62{}^{20}{C}_{20}$

$$\begin{gathered} =\sum _{r=0}^{20}(3r+2){}^{20}{C}_r \\ =\underset{r=0}{\overset{20}{3\sum \left(r\right)·}}{}^{20}{C}_r+2\sum _{r=0}^{20}{}^{20}{C}_r \\ =\underset{r=0}{\overset{20}{3\sum \left(r\right)·}}\frac{20!}{\left(r\right)!(20-r)!}+2\left({}^{20}{C}_0+{}^{20}{C}_1+........+{}^{20}{C}_{20}\right) \\ =\underset{r=0}{\overset{20}{3\sum \left(r\right)·}}\frac{20\left(19!\right)}{\left(r\right)(r-1)!(19-(r-1\left)\right)!}+2\left({}^{20}{C}_0+{}^{20}{C}_1+........+{}^{20}{C}_{20}\right) \\ =3\times 20\times \sum _{r=0}^{20}\frac{\left(19!\right)}{(r-1)!(19-(r-1\left)\right)!}+2\left({}^{20}{C}_1+{}^{20}{C}_2+........+{}^{20}{C}_{20}\right) \\ =60\sum _{r=1}^{20}{}^{19}{C}_{r-1}+2\times 2^{20}\left[{}^{\mathbf{n}}{\mathbf{C}}_{\mathbf{1}}\mathbf{+}{}^{\mathbf{n}}{\mathbf{C}}_{\mathbf{2}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{+}{}^{\mathbf{n}}{\mathbf{C}}_{\mathbf{n}}\mathbf{=}{\mathbf{2}}^{\mathbf{n}}\right] \\ =60\times 2^{19}+2\times 2^{20} \\ =15\times 2^{21}+2^{21} \\ =2^{21}\left(15+1\right) \\ =2^{21}\left(16\right) \end{gathered}$$

Hence, option D is correct.