\(\text { Ans. } x=\frac{2}{3}+i \frac{\sqrt{14}}{21}\)

We know that,
\(\left(x_{1}+x_{2}+\ldots+x_{n}\right)^{2}=\left(x_{1}^{2}+x_{2}^{2}+\ldots+x_{n}^{2}\right)\)
+ 2(Sum of all possible products taken two at a time)
\(\text { or }\left(\sum_{i=1}^{n} x_{i}\right)^{2}=\left(\sum_{i=1}^{n} x_{i}^{2}\right)+2\left(\sum_{i=1, i<j}^{n} \sum_{j=1}^{n} x_{i} x_{j}\right) \)
\(\Rightarrow \sum_{i=1, i<j}^{n} \sum_{j=1}^{n} x_{i} x_{j}=\frac{1}{2}\left\{\left(\sum_{i=1}^{n} x_{i}\right)^{2}-\left(\sum_{j=1}^{n} x_{i}^{2}\right)\right\}\)
\(\therefore\) Required sum \(=\frac{1}{2}\left\{\left(\sum_{k=1}^{n} k\right)^{2}-\left(\sum_{k=1}^{n} k^{2}\right)\right\} \)
\(=\frac{1}{2}\left[\left\{\frac{n(n+1)}{2}\right\}^{2}-\frac{n(n+1)(2 n+1)}{6}\right] \)
\(=\frac{1}{2}\left[\frac{n(n+1)}{2}\left\{\frac{n(n+1)}{2}-\frac{2 n+1}{3}\right\}\right] \)
\(=\frac{n(n+1)}{4}\left\{\frac{3 n^{2}+3 n-4 n-2}{6}\right\}\)
\(=\frac{n(n+1)\left(3 n^{2}-n-2\right)}{24}=\frac{n(n+1)(n-1)(3 n+2)}{24}\)
 

As 2n is even so the expansion (1 + x)²ⁿ has only one middle term which is
\(\left(\frac{2 n}{2}+1\right)^{\text {th }}\) i.e., (n + 1)^th term.
The (n + 1)^th term is ²ⁿC_nxⁿ. The coefficient of xⁿ is ²ⁿC_n
Similarly, (2n – 1) being odd, the other expansion has two middle terms 
\(\left(\frac{2 n-1+1}{2}\right)^{\text {th }} \text { and }\left(\frac{2 n-1+1}{2}+1\right)^{\text {th }}\) i.e., n^th and (n + 1)^th terms. The coefficients of these terms are ^{2n – 1}C_{n – 1} and ^{2n – 1}C_n, respectively.
Now \({ }^{2 n-1} \mathrm{C}_{n-1}+{ }^{2 n-1} \mathrm{C}_{n}={ }^{2 n} \mathrm{C}_{n}\) [ As \({ }^{n} \mathrm{C}_{r-1}+{ }^{n} \mathrm{C}_{r}={ }^{n+1} \mathrm{C}_{r}\)] as required.

Ans. (5 ,\(\infty\))

Let 1 = r cos \(\theta\), \(\sqrt{3}\)  = r sin \(\theta\)
By squaring and adding, we get r² (cos²\(\theta\)+ sin²\(\theta\)) = 4
i.e., r = \(\sqrt{4}\) = 2 (conventionally, r >0)
Therefore, cos \(\theta\) \(=\frac{1}{2}\), sin \(\theta\) \(=\frac{\sqrt{3}}{2}\), which gives \(\theta\) \(=\frac{\pi}{3}\)
Therefore, required polar form is \(z=2\left(\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}\right)\)
The complex number z = 1+ i\(\sqrt{3}\)  is represented.

The graphical representation of – 3x + 2y = – 6 is given in the figure below.
This line divides the xy-plane in two half planes.
Select a point (not on the line), which lies in one of the half planes, to determine whether the point satisfies the given inequality or not.
We select the point as (0, 0).
It is observed that,
– 3(0) + 2(0) ≥ – 6 or 0 ≥ –6, which is true
Therefore, the lower half plane is not the solution region of the given inequality. Also, it is evident that any point on the line satisfies the given inequality.
Thus, the solution region of the given inequality is the half plane containing the point (0, 0) including the line.
The solution region is represented by the shaded region as follows.

(i) Total number of ways of selecting 11players out of 17 is
\(^{17}C_{11}={17!\over 11!6!}={17\times16\times15\times14\times13\times12\times11!\over 11!\times6\times5\times4\times3\times2\times1}=12376\)
(ii) A particular player is always chosen means that 10 players are selected out of remaining 16 players is
\(^{16}C_{10}={16!\over 10!6!}={16\times15\times14\times13\times12\times11\times10!\over 10!\times6\times5\times4\times3\times2\times1}=8008\)
(iii)A particular player is never chosen means that 11 players are selected out of remaining 16 players is
\(^{16}C_{11}={16!\over 11!5!}={16\times15\times14\times13\times12\times11!\over 11!\times6\times5\times4\times3\times2\times1}=4368\)

\(\frac { a }{ r } +a+ar=52\Rightarrow a\left( \frac { 1 }{ r } +1+r \right) =52\) ...(i)
and \(\frac { a }{ r } .a+a.ar+\frac { a }{ r } .ar\Rightarrow 624\Rightarrow { a }^{ 2 }\left( \frac { 1 }{ r } +r+1 \right) =624\) .(ii)
On dividing Eq.(ii) by Eq(i) we get a=12
On putting a=12 in Eq(i) we get
\(12\left( \frac { 1 }{ r } +r+1 \right) =52\Rightarrow r=\frac { 1 }{ 3 } \) or r=3
Ans.When \(r=\frac { 1 }{ 3 } \),then number are 36,12,44
When r=3, then numbers are 4, 12, 36.

\(\left( c+dy \right) ^{ -2 }={ c }^{ -2 }(1+\frac { d }{ c } y)^{ -2 }\)
\(\therefore c^{ -2 }[1-2\times \frac { d }{ c } y+\frac { \left( -2 \right) \left( -2-1 \right) }{ 2! } \left( \frac { d }{ c } y \right) ^{ 2 }+...]=\frac { 1 }{ 4 } -3y+...\)
\(\Rightarrow { c }^{ -2 }-\frac { 2d }{ { c }^{ 3 } } y+...=\frac { 1 }{ 4 } -3y+...\)
\(On\quad comparing\quad first\quad term,{ c }^{ -2 }=\frac { 1 }{ 4 } \Rightarrow \frac { 1 }{ { c }^{ 2 } } =\frac { 1 }{ 4 } \Rightarrow c=2\)
\(On\quad comparing\quad second\quad term,\quad d=12\)
\( Ans.2,12\)

n=6

(i) (a)
(ii) (c)
(iii) (b)
(iv) (a)
(v) (b)