Question

If $(1+x{)}^n=C_0+C_{1}x+C_2{x}^2+.......+C_n{x}^n$, then

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

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

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

= $\frac{n}{1}+2\frac{\frac{n(n-1)}{1.2}}{n}+3\frac{\frac{n(n-1)(n-2)}{\mathrm{3.2.1}}}{\frac{n(n-1)}{1.2}}+......+n.\frac{1}{n}$

=n + (n-1) + (n-2)........ +1 = $\sum$n = $\frac{n(n+1)}{2}$

Trick: Put n = 1, 2, 3........., then $S_1$ = $\frac{{}^1{C}_1}{{}^1{C}_0}$ = 1,

$S_2$ = $\frac{{}^2{C}_1}{{}^2{C}_0}$ + 2 $\frac{{}^2{C}_2}{{}^2{C}_1}$ = $\frac{2}{1}$ + 2. $\frac{1}{2}$ = 2 + 1 = 3

By option, (put n=1, 2..........) (a) and (b) does not

hold condition, but (c) $\frac{n(n+1)}{2}$, put n = 1, 2.........

$S_1$ = 1, $S_2$ = 3 which is correct.