Question

Find $\frac{C_1}{C_0}+\frac{2C_2}{C_1}+\frac{3C_3}{C_2}+....+\frac{n{C}_n}{C_{n-1}}=$

Answer 1

$S=\frac{{}^n{C}_1}{{}^n{C}_0}+2\frac{{}^n{C}_2}{{}^n{C}_1}+3\frac{{}^n{C}_3}{{}^n{C}_2}+......+n\frac{{}^n{C}_n}{{}^n{C}_{n-1}}$

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

Now,

$\frac{{}^n{C}_r}{{}^n{C}_{r-1}}=\frac{\frac{n!}{r!(n-r)!}}{\frac{n!}{(r-1)!(n-r+1)!}}$

$=\frac{(r-1)!(n-r+1)!}{r!(n-r)!}$

$=\frac{(r-1)!(n-r+1)(n-r)!}{r(r-1)!(n-r)!}$

$=\frac{n-r+1}{r}$

Hence (1) become

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

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

$={\sum }_{r=1}^{n}n+{\sum }_{r=1}^{n}1-{\sum }_{r=1}^{n}r$

$=n(1+1+\mathrm{1.....}n)+(1+1+\mathrm{1.....}n)-(1+2+3+....+n)$

$=n\left(n\right)+n-\frac{n(n+1)}{2}$

$=(n^2+n)-\frac{n^2+n}{2}$

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

$\therefore \frac{{}^n{C}_1}{{}^n{C}_0}+2\frac{{}^n{C}_2}{{}^n{C}_1}+3\frac{{}^n{C}_3}{{}^n{C}_2}+......+n\frac{{}^n{C}_n}{{}^n{C}_{n-1}}=\frac{1}{2}(n^2+n)$