Free Objective Test 02 Practice Test - 11th and 12th 

= $(1+3x{)}^2$($1-2x{)}^{-1}$

 = ($1+3x{)}^2$(1 + 2x + $\frac{1.2}{2.1}(-2x{)}^2+.......$)

= $(1+6x+9x^2)(1+2x+4x^2+8x^3+........)$

Therefore coefficient of $x^3$ is (8 + 24 + 18) = 50.

The expansion is $x^{-\frac{8}{3}}(1+\frac{1}{x^3}{)}^{-}\frac{4}{3}$

Hence | $\frac{1}{x^3}$| < 1  ⇒ |x| > 1

Given term can be written as $(1+x{)}^2$$(1-x{)}^{-2}$

=$(1+2x+x^2)[1+2x+3x^2+.......+(n-1)x^{n-2}+n{x}^{n-1}+(n+1)x^n+...........]$

= $x^n(n+1+2n+n-1)$+..........

Therefore coefficient of $x^n$ is 4n.

$(1-x{)}^{-4}$ = [$\frac{\mathrm{1.2.3}}{6}{x}^0+\frac{\mathrm{2.3.4}}{6}x+\frac{\mathrm{3.4.5}}{6}{x}^2+\frac{\mathrm{4.5.6}}{6}{x}^3+.......+\frac{(r+1)(r+2)(r+3)}{6}{x}^r+.........]$

Therefore $T_{r+1}$ =  $\frac{(r+1)(r+2)(r+3)}{6}{x}^r$.

$(\frac{a+x}{a}{)}^{\frac{-1}{2}}$ +  $(\frac{a-x}{a}{)}^{\frac{-1}{2}}$ =$(1+\frac{x}{a}{)}^{\frac{-1}{2}}+(1-\frac{x}{a}{)}^{\frac{-1}{2}}$

=[1 + (- $\frac{1}{2}$)( $\frac{x}{a}$) + $\frac{(-\frac{1}{2})(-\frac{3}{2})}{2.1}$$(\frac{x}{a}{)}^2$ + ........]

+ $[1+(-\frac{1}{2}\left)\right(-\frac{x}{a})+\frac{(-\frac{1}{2})\left(\frac{3}{2}\right)}{2.1}(-\frac{x}{a}{)}^2+........]$

= 2 +  $\frac{3x^2}{4a^2}$ + ......

Here odd terms cancel each other.

$(217{)}^{\frac{1}{3}}$ = $(6^3+1{)}^{\frac{1}{3}}$ = $6(1+\frac{1}{6^3}{)}^{\frac{1}{3}}$

On expansion by binomial theorem

= $6(1+\frac{1}{3\times 216}-\frac{1\times 2}{3\times 3\times 2}(\frac{1}{216}{)}^2+.........)$ = 6.01

Expansion of $(1-2x{)}^{\frac{3}{2}}$

= $1+\frac{3}{2}(-2x)+\frac{3}{2}.\frac{1}{2}.\frac{1}{2}(-2x{)}^2+\frac{3}{2}.\frac{1}{2}(-\frac{1}{2}\left)\frac{1}{6}\right(-2x{)}^3+.........$

Hence $4^{th}$ term is  $\frac{x^3}{2}$

We know that

$(1+x{)}^n{=}^n{C}_0{+}^n{C}_{1}x{+}^n{C}_2{x}^2+......{+}^n{C}_n{x}^n$

Putting x = -1, we get

$(1-1{)}^n$ = ${}^n{C}_0$ -  ${}^n{C}_1$ +  ${}^n{C}_2$ - .......$(-1{)}^n$  ${}^n{C}_n$

Therefore $C_0-C_1+C_2-C_3+......(-1{)}^n{C}_n$ = 0

We know that 

$\frac{(1+x{)}^n-(1-x{)}^n}{2}$ = $C_{1}x+C_3{x}^3+C_5{x}^5+........$

Integrating from x = 0 to x = 1, we get

$\frac{1}{2}$${\int }_0^1(1+x{)}^n-(1-x{)}^{n}dx$

= ${\int }_0^1(C_{1}x+C_3{x}^3+C_5{x}^5+.......)dx$

⇒ $\frac{1}{2}{\{\frac{(1+x{)}^{n+1}}{n+1}+\frac{(1-x{)}^{n+1}}{n+1}\}}_0^1=\frac{C_1}{2}$ +  $\frac{C_3}{4}$ + $\frac{C_5}{6}$ + ....

or  $\frac{C_1}{2}$ +  $\frac{C_3}{4}$ + $\frac{C_5}{6}$ + .......=  $\frac{1}{2}$  {$\frac{2^{n+1}-1}{n+1}$ +  $\frac{0-1}{n+1}$}

=  $\frac{1}{2}$ $\frac{2^{n+1}-2}{n+1}$ =  $\frac{2^n-1}{n+1}$

We can write

a$C_0$ - (a + d)$C_1$ + (a + 2d)$C_2$ - ........ upto (n+1) terms

$=a(C_0-C_1+C_2-........)+d(-C_1+2C_2-3C_3+......)$            .............(i)

Again,$(1-x{)}^n=C_0-C_{1}x+C_2{x}^2-.........+(-1{)}^n{C}_n{x}^n$   ..........(ii)

Differentiating with respect to x,

$-n(1-x{)}^{n-1}$ = -$C_1$ + 2$C_2$x - .......... + $(-1{)}^n{C}_{n}n{x}^{n-1}$     ......................(iii)

Putting x = 1 in (ii) and (iii), we get

$C_0$ - $C_1$ + $C_2$ - ........ + (-1)${}^n{C}_n$ = 0

and -$C_1$ + 2$C_2$ - ........+$(-1{)}^{n}n.C_n$ = 0

Thus the required sum to (n+1) terms, by (i)

= a.0 + d.0 = 0.

Putting the values of $C_0,C_2,C_4$........., we get

= 1 +  $\frac{n(n-1)}{3.2!}$ +  $\frac{n(n-1)(n-2)(n-3)}{5.4!}$ + .........

= $\frac{1}{n+1}$[(n+1) + $\frac{(n+1)n(n-1)}{3!}$ +  $\frac{(n+1)n(n-1)(n-2)(n-3)}{5!}$ + ..........]

Put n+1 = N

=  $\frac{1}{N}$[N +  $\frac{N(N-1)(N-2)}{3!}$ +  $\frac{N(N-1)(N-2)(N-3)(N-4)}{5!}$ + .........]

=  $\frac{1}{N}${${}^N{C}_1$ + ${}^N{C}_3$ + ${}^N{C}_5$ +..........}

= $\frac{1}{N}${$2^{N-1}$} =  $\frac{2^n}{n+1}$      { ∵ N = n+1}

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

At n = 2, $S_2$ =   $\frac{{}^2{C}_0}{1}$ +   $\frac{{}^2{C}_2}{3}$ = 1 +   $\frac{1}{3}$ =   $\frac{4}{3}$

Also (c)  ⇒ $S_1$ = 1, $S_2$ =  $\frac{4}{3}$

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

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

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

$(1+x{)}^n=C_0+C_{1}x+C_2{x}^2+.........+C_n{X}^n$  ...........(i)

and $(1+\frac{1}{x}{)}^n$ =$C_0+C_1\frac{1}{x}+C_2(\frac{1}{x}{)}^2+........+C_n(\frac{1}{x}{)}^n$    ..........(ii)

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

$C_0^2+C_1^2+C_2^2+........+C_n^2$

is the term independent of x and hence it is equal to the term independent of x in the product $(1+x{)}^n(1+\frac{1}{x}{)}^n$ or in  $\frac{1}{x^n}$$(1+x{)}^{2n}$ or term

containing $x^n$ in $(1+x{)}^{2n}$. Clearly the coefficient of $x^n$ in $(1+x{)}^{2x}$ is $T_{n+1}$ and equal to ${}^{2n}{C}_n$ = $\frac{\left(2n\right)!}{n!n!}$ 

Trick: Solving conversely.

Put n = 1, n = 2,.........then we get $S_1$ = ${}^1{C}_0^2{+}^1{C}_1^2$= 2,

$S_2$ = ${}^2{C}_0^2{+}^2{C}_1^2{+}^2{C}_2^2=1^2+2^2+1^2$ = 6

Now check  the options
(a) Does not hold given condition,

(b) (i) Put n = 1, then $\frac{2!}{1!1!}=2$
     (ii) Put n = 2, then $\frac{4!}{2!2!}=\frac{4\times 3\times 2\times 1}{2\times 1\times 2\times 1}=6$

Note: Students should remember this question as identity

$T_{r+1}={12}_{c_r}{\left(\frac{a}{x}\right)}^{12-r}{\left(bx\right)}^r={12}_{c_r}.a^{12-r}.b^r.x^{2r-12}\therefore 2x-12=-10\Rightarrow r=1\therefore \text{coefficient}=12.a^{11}.b$

2${}^n{c}_p=n_{c_{p-1}}+p_{c_{p+1}}\Rightarrow \frac{2}{(n-p)!P!}=\frac{1}{(n-p+1)!(p-1)!}+\frac{1}{(n-p-1)!(p+1)!}\Rightarrow \frac{2}{(n-p)p}=\frac{1}{(n-p+1)(n-p)}+\frac{1}{(p+1)p}onsimplifingweget{n}^2-n(4p+1)+4p^2-2=0$