Question

$x^{n}$=

• A. $A = B$
• B. $A = 2B$
• C. $2A = B$
• D. None of these

Answer 1

Answer: (A)

$A = B$

Answer 2

$\frac{n!}{n!n!}$ …..(i)

$\frac{(2n)!}{n!n!}$ …..(ii)

Multiplying both sides and equating coefficient of $\frac{(2n)!}{n!}$in $(1 + x)^{n} = C_{0} + C_{1}x + C_{2}x^{2} + .......... + C_{n}x^{n}$or the coefficient of $\frac{C_{1}}{C_{0}} + \frac{2C_{2}}{C_{1}} + \frac{3C_{3}}{C_{2}} + .... + \frac{nC_{n}}{C_{n - 1}} =$in $\frac{n(n - 1)}{2}$ we get the value of required expression = $\frac{n(n + 2)}{2}$

Trick : Solving conversely.

Put $C_{1} + 2C_{2} + 3C_{3} + 4C_{4} + .... + nC_{n} =$ and $2^{n}$ in first term , (given condition)

(i) $n.2^{n + 1}$ , $\frac{C_{0}}{1} + \frac{C_{2}}{3} + \frac{C_{4}}{5} + \frac{C_{6}}{7} + ....$

Put $\frac{2^{n + 1}}{n + 1}$, then

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

Now check the options

(1) (i) Put $\frac{2^{n}}{n + 1}$, we get $\frac{C_{0}}{1} + \frac{C_{1}}{2} + \frac{C_{2}}{3} + .... + \frac{C_{n}}{n + 1} =$

(ii) Put $\frac{2^{n}}{n + 1}$, we get $\frac{2^{n} - 1}{n + 1}$

Note : Students should remember this question as an identity.