Question

If ${}^n{C}_r=\frac{n!}{r!(n-r)!}$ then the sum of the series $1+{}^n{C}_r+{}^n+1C_r+{}^{n+2}{C}_3+.........+{}^{n+r-1}{C}_r$ is

• A. ${}^{n+r}{C}_r$
• B. ${}^{n+r-1}{C}_r$
• C. ${}^{n+r}{C}_r-1$
• D. none of these

Answer 1

Answer: (A)

${}^{n+r}{C}_r$

Given that:

$S=1{+}^n{c}_1{+}^{n+1}{c}_2{+}^{n+2}{c}_3+.....{+}^{n+r-1}{c}_r$

$S{=}^n{c}_n{+}^n{c}_1{+}^{n+1}{c}_2{+}^{n+2}{c}_3+.....{+}^{n+r-1}{c}_r$

Now, using the property ${}^n{c}_r{=}^n{c}_{n-r}$

$S{=}^n{c}_n{+}^n{c}_{n-1}{+}^{n+1}{c}_{n-1}{+}^{n+2}{c}_{n-1}+.....{+}^{n+r-1}{c}_{n-1}$

$S{=}^{n+1}{c}_n{+}^{n+1}{c}_{n-1}{+}^{n+2}{c}_{n-1}+.....{+}^{n+r-1}{c}_{n-1}$

$S{=}^{n+2}{c}_n{+}^{n+2}{c}_{n-1}+.....{+}^{n+r-2}{c}_{n-1}{+}^{n+r-1}{c}_{n-1}$

$...$

$....$

$S{=}^{n+r-1}{c}_n{+}^{n+r-1}{c}_{n-1}$

$S{=}^{n+r}{c}_n$

Again using the property ${}^n{c}_r{=}^n{c}_{n-r}$, we get

$S={}^{n+r}{c}_r$.