Question

$\sum _{r=0}^m$ ${}^{n+r}{C}_n$ is equal to

• A. ${}^{n+m+1}{C}_{n+1}$
• B. ${}^{n+m+2}{C}_n$
• C. ${}^{n+m+3}{C}_{n-1}$
• D. None of these.

Answer 1

The correct option is A ${}^{n+m+1}{C}_{n+1}$

Since ${}^n{C}_r{=}^n{C}_{n-r}$

and ${}^n{C}_{r-1}{+}^n{C}_r{=}^{n+1}{C}_r$, we have

$\sum _{r=0}^m$ ${}^{n+r}{C}_n$ = $\sum _{r=0}^m$ ${}^{n+r}{C}_r$ = ${}^n{C}_0{+}^{n+1}{C}_1{+}^{n+2}{C}_2+.......{+}^{n+m}{C}_m$

=$[1+(n+1\left)\right]{+}^{n+2}{C}_2{+}^{n+3}{C}_3+.......{+}^{n+m}{C}_m$

= ${(}^{n+1}{C}_1{+}^{n+2}{C}_2){+}^{n+3}{C}_3+.......{+}^{n+m}{C}_m$

= $\because n+2{=}^{n+1}{C}_1$ or ${}^n{C}_1=n$

= ${(}^{n+3}{C}_2{+}^{n+3}{C}_3)+.......{+}^{n+m}{C}_m$

= ${(}^{n+4}{C}_3{+}^{n+4}{C}_4)+.......{+}^{n+m}{C}_m$

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= ${}^{n+m}{C}_{m-1}{+}^{n+m}{C}_m$

=${}^{n+m+1}{C}_m{=}^{n+m+1}{C}_{n+1}$ $\left[{\because }^n{C}_r{=}^n{C}_{n-r}\right]$