Question

For $2\le r\le n$, $\left(\begin{array}{c}n\\ r\end{array}\right)+2\left(\begin{array}{c}n\\ r-1\end{array}\right)+\left(\begin{array}{c}n\\ r-2\end{array}\right)=$

• A. $\left(\begin{array}{c}n+1\\ r-1\end{array}\right)$
• B. $2\left(\begin{array}{c}n+1\\ r+1\end{array}\right)$
• C. $2\left(\begin{array}{c}n+2\\ r\end{array}\right)$
• D. $\left(\begin{array}{c}n+2\\ r\end{array}\right)$

Answer 1

The correct option is D $\left(\begin{array}{c}n+2\\ r\end{array}\right)$

${=}^n{C}_r+{2.}^n{C}_{r-1}{+}^n{C}_{r-2}$
${=}^n{C}_r+\frac{2r}{(n-r+1)}{.}^n{C}_r+\frac{r(r-1)}{(n-r+1)(n-r+2)}{.}^n{C}_r$
${=}^n{C}_r(\frac{n+r+1}{(n-r+1)}+\frac{r(r-1)}{(n-r+1)(n-r+2)})$
${=}^n{C}_r\left(\frac{n^2+3n+2}{(n-r+1)(n-r+2)}\right)$
${=}^n{C}_r\left(\frac{(n+1)(n+2)}{(n-r+1)(n-r+2)}\right)$
${=}^{n+2}{C}_r$
Hence, option 'D' is correct.