Question

If $c_0,c_1,c_2,....c_n$ denote the coefficients in the expansion of $(1+x{)}^n$, prove that $\frac{c_1}{c_0}+\frac{2c_2}{c_1}+\frac{3c_3}{c_2}+.....+\frac{n{c}_n}{c_{n-1}}=\frac{n(n+1)}{2}$.

Answer 1

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

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

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 ) becomes,

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

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

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

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

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

$=n\left(n\right)+n-\frac{n(n+1)}{2}$ [ Sum of $n$ natural number $=\frac{n(n+1)}{2}$ ]

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

$=\frac{n^2+n}{2}$

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