Question

(C₁ / C₀) + 2 (C₂ / C₁) + 3 (C₃ / C₂) + …. + n (C_n / C_n-1) =

• A. $\frac{n(n-1)}{2}$
• B. $\frac{n(n+1)}{2}$
• C. $\frac{(n+1)(n+2)}{2}$
• D. None of these

Answer 1

The correct option is B

$\frac{n(n+1)}{2}$

Explanation for correct option:

Step 1: Declaring the theorem

We know, ${}^n{C}_r=\frac{n!}{(n-r)!r!}$

$n!=n(n-1)!$

Step 2: Solving the given equation

$\frac{C_1}{C_0}+2\frac{C_2}{C_1}+3\frac{C_3}{C_2}+....+n\frac{C_n}{C_{n-1}}$

Therefore, S = $\sum _{r=1}^n\left(r\right)\frac{{}^n{C}_r}{{}^n{C}_{r-1}}$

$\Rightarrow$ S = $\sum _{r=1}^n\left(r\right)\frac{{\displaystyle \frac{n!}{(n-r)!r!}}}{{\displaystyle \frac{n!}{(n-r+1)!(r-1)!}}}$

$\Rightarrow$S = $\sum _{r=1}^n\left(r\right)\frac{(n-r+1)!(r-1)!}{(n-r)!r!}$

$\Rightarrow$ S = $\sum _{r=1}^n\left(r\right)\frac{(n-r+1)!(n-r)!(r-1)!}{(n-r)!r(r-1)!}$

$\Rightarrow$ S = $\sum _{r=1}^n\left(r\right)\frac{n-r+1}{r}$

$\Rightarrow$ S = $\sum _{r=1}^n(n+1)-\sum _{r=1}^{n}r$

$\Rightarrow$ S = $(n+1)\underset{r=1}{\overset{n}{\sum \left(1\right)}}-\sum _{r=1}^{n}r$

$\Rightarrow$ S = $(n+1)n-\{1+2+3+4....n\}$

$\Rightarrow$ S = $(n+1)n-\frac{n(n+1)}{2}$

$\Rightarrow$ S = $\frac{n(n+1)}{2}$

Therefore, we can prove that $\frac{C_1}{C_0}+2\frac{C_2}{C_1}+3\frac{C_3}{C_2}+....+n\frac{C_n}{C_{n-1}}$ $=\frac{n(n+1)}{2}$

Hence, Option(B) is correct.