Question

${}^{m}C_{r+1}+\sum _{k=m}^n{}^{k}C_r=$

• A. ${}^{n}C_{r+1}$
• B. ${}^{n+1}C_{r+1}$
• C. ${}^{n}C_r$
• D. None of these

Answer 1

The correct option is B

${}^{n+1}C_{r+1}$

Explanation for the correct option:

Finding the value of ${}^{m}C_{r+1}+\sum _{k=m}^n{}^{k}C_r$ :

Formula to be used : According to the binomial theorem, ${}^m{C}_r{+}^m{C}_{r+1}{=}^{m+1}{C}_{r+1}$.

Then,

${}^{\mathrm{m}}{C}_{\mathrm{r}+1}+\sum _{k=m}^n{}^{\mathrm{k}}C_{\mathrm{r}}{=}^m{C}_{r+1}+\left[{}^m{C}_r{+}^{m+1}{C}_r{+}^{m+2}{C}_r+...{+}^n{C}_{r+1}\right]$

$$\begin{gathered} =\left({}^m{C}_{r+1}{+}^m{C}_r\right){+}^{m+1}{C}_r{+}^{m+2}{C}_r+...{+}^n{C}_r\mathbf{[}\mathbf{\because }{}^{\mathbf{n}}{\mathbf{C}}_{\mathbf{r}}\mathbf{+}{}^{\mathbf{n}}{\mathbf{C}}_{\mathbf{r}\mathbf{-}\mathbf{1}}\mathbf{=}{}^{\mathbf{n}\mathbf{+}\mathbf{1}}{\mathbf{C}}_{\mathbf{r}}\mathbf{]} \\ {=}^{m+1}{C}_{r+1}{+}^{m+1}{C}_r{+}^{m+2}{C}_r+...{+}^n{C}_r \\ =\left({}^{m+1}{C}_{r+1}{+}^{m+1}{C}_r\right){+}^{m+2}{C}_r+...{+}^n{C}_r \\ {=}^{m+2}{C}_{r+1}{+}^{m+2}{C}_r+...{+}^n{C}_r \\ ⋮ \\ {=}^n{C}_{r+1}{+}^n{C}_r \\ {=}^{n+1}{C}_{r+1} \end{gathered}$$

Hence, option (B) is the correct answer.