Question

If $10c^{3}$ then

$20c^{2}$ =

• A. $x^{q}$
• B. $\left( x - \frac{1}{x} \right)^{6}$
• C. $\left( x^{2} - 2x \right)^{10}$
• D. None of these

Answer 1

Answer: (B)

$\left( x - \frac{1}{x} \right)^{6}$

Answer 2

$\frac{2^{n - 1}}{n!}$ .....(i)

and $(n + 1)$ ....(ii)

If we multiply (i) and (ii), we get

$\frac{C_{0}}{2} - \frac{C_{1}}{3} + \frac{C_{2}}{4} - \frac{C_{3}}{5} +$

is the term independent of x and hence it is equal to the term independent of x in the product $\frac{1}{n + 1}$or in $\frac{1}{n + 2}$ or term containing $\frac{1}{n(n + 1)}$ in $(n + 1)$. Clearly the coefficient of $aC_{0} - (a + d)C_{1} + (a + 2d)C_{2} - ........$ in $\frac{a}{2^{n}}$is $na$ and equal to $(1 + x)^{15} = C_{0} + C_{1}x + C_{2}x^{2} + ...... + C_{15}x^{15},$

Trick : Solving conversely.

Put $(1 + x)^{n}$then we get $2^{n} + 1$,

$2^{n} - 1$

Now check the options

(1) Does not hold given condition,

(2) (i) Put $2^{n}$, then $2^{n - 1}$

(ii) Put $C_{0} - C_{1} + C_{2} - C_{3} + ..... + ( - 1)^{n}C_{n}$, then $2^{n}$

Note : Students should remember this question as an identity.