Question

For $r=0,1,...,10$let $A_r,B_r$and $C_r$ denote, respectively the coefficient of $x_r$ in the expansion of${(1+x)}^{10},{(1+x)}^{20}$ and ${(1+x)}^{30}.$

Then $\sum _{r=1}^{10}{A}_r(B_{10}{B}_r-C_{10}{A}_r)=?$

• A. $B_{10}–C_{10}$
• B. $A_{10}({B_{10}}^2–C_{10}{A}_{10})$
• C. $0$
• D. $C_{10}–B_{10}$

Answer 1

The correct option is D

$C_{10}–B_{10}$

The explanation for the correct option:

Step1. Calculate the coefficient of ${\mathit{A}}_{\mathbf{r}\mathbf{}\mathbf{,}}{\mathit{B}}_{\mathbf{r}\mathbf{,}\mathbf{}}{\mathit{C}}_{\mathbf{r}\mathbf{:}}$

Using the binomial theorem we know,

$A_r=$ coefficient of $x_r$ in ${\left(1+\mathrm{x}\right)}^{10}={}^{10}C_r$

$B_r=$ coefficient of $x_r$ in ${\left(1+\mathrm{x}\right)}^{20}={}^{20}C_r$

$C_r=$ coefficient of $x_r$ in ${\left(1+\mathrm{x}\right)}^{30}={}^{30}C_r$

Step2. Calculate the value of $\underset{\mathbf{r}\mathbf{=}\mathbf{1}}{\overset{\mathbf{10}}{\mathbf{\sum }{\mathbf{A}}_{\mathbf{r}}\mathbf{}}}{\mathbf{B}}_{\mathbf{r}}\mathbf{}$ and $\underset{\mathbf{r}\mathbf{=}\mathbf{1}}{\overset{\mathbf{10}}{\mathbf{\sum }{\mathbf{A}}_{\mathbf{r}}\mathbf{}}}{\mathbf{A}}_{\mathbf{r}}$.

$\underset{\mathrm{r}=1}{\overset{10}{\sum {\mathrm{A}}_{\mathrm{r}}}}{\mathrm{B}}_{\mathrm{r}}=$coefficient of $\begin{array}{rcl}& & \left[{(1+\mathrm{x})}^{10}{\left(1+\mathrm{x}\right)}^{20}\right]-1\end{array}$

$=$coefficient of $\begin{array}{rcl}& & \left[{(1+\mathrm{x})}^{10+20}\right]-1\end{array}$

$=$coefficient of $\begin{array}{l}\left[{(1+\mathrm{x})}^{30}\right]-1\end{array}$

$\begin{array}{rcl}& =& {}^{30}{C}_r-1\mathbf{[}\mathbf{\because }{}^{\mathbf{30}}{\mathbf{C}}_{\mathbf{r}}\mathbf{}\mathbf{=}{\mathbf{C}}_{\mathbf{r}}\mathbf{]}\\ & =& {\mathrm{C}}_{\mathrm{r}}-1\left[\mathbf{\because }\mathbf{r}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{}\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{,}\mathbf{3}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}\mathbf{10}\right]\\ & =& {\mathrm{C}}_{10}-1.....\left(i\right)\end{array}$

$\underset{\mathrm{r}=1}{\overset{10}{\sum {\mathrm{A}}_{\mathrm{r}}}}{A}_{\mathrm{r}}=$coefficient of $\begin{array}{l}\left[{(1+\mathrm{x})}^{10}{\left(1+\mathrm{x}\right)}^{20}\right]-1\end{array}$

$=$coefficient of $\begin{array}{l}\left[{(1+\mathrm{x})}^{10+20}\right]-1\end{array}$

$=$coefficient of $\begin{array}{l}\left[{(1+\mathrm{x})}^{30}\right]-1\end{array}$

$\begin{array}{rcl}& =& {}^{20}{C}_r-1\mathbf{[}\mathbf{\because }{}^{\mathbf{20}}{\mathbf{C}}_{\mathbf{r}}\mathbf{}\mathbf{=}{\mathit{B}}_{\mathbf{r}}\mathbf{]}\\ & =& B_{\mathrm{r}}-1\left[\mathbf{\because }\mathbf{r}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{}\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{,}\mathbf{3}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}\mathbf{10}\right]\\ & =& B_{10}-1.....\left(ii\right)\end{array}$

Step3.Calculate the value of $\mathbf{\sum }_{\mathbf{r}\mathbf{=}\mathbf{1}}^{\mathbf{10}}\mathbf{}{\mathit{A}}_{\mathbf{r}}\mathbf{}\mathbf{(}{\mathit{B}}_{\mathbf{10}\mathbf{}}{\mathit{B}}_{\mathbf{r}}\mathbf{}\mathbf{-}\mathbf{}{\mathit{C}}_{\mathbf{10}}\mathbf{}{\mathit{A}}_{\mathbf{r}}\mathbf{)}\mathbf{}$:

$$\begin{gathered} \sum _{r=1}^{10}{A}_r(B_{10}{B}_r-C_{10}{A}_r) \\ =\sum _{r=1}^{10}{B}_{10}{A}_r{B}_r-\sum _{r=1}^{10}{A}_r{C}_{10}{A}_r) \end{gathered}$$

$=B_{10}\sum _{r=1}^{10}{A}_r{B}_r-C_{10}\sum _{r=1}^{10}{A}_r{A}_r$ (from equation (i) and (ii))

$$\begin{gathered} =B_{10}\left(C_{10}-1\right)-C_{10}\left(B_{10}-1\right) \\ =B_{10}{C}_{10}-B_{10}-C_{10}{B}_{10}+C_{10} \\ ={\mathit{C}}_{\mathbf{10}}\mathbf{-}{\mathit{B}}_{\mathbf{10}} \end{gathered}$$

Hence, Option(D) is the correct answer.