Binomial Theorem

= A + B (say)

(1 + x)¹⁵ =

differentiating 15 (1 + x)¹⁴ =   $\sum r{.}^{15}{C}_r{x}^{r-1}$

put  = x = -1

$\Rightarrow 0={-}^{15}{C}_1+2{.}^{15}{C}_2-.......=A$      

$B{=}^{14}{C}_{13}=2^{13}$    

$\therefore A+B=2^{13}-14$     

${\displaystyle \sum _{i=0}^k\left(\begin{array}{l}10\\ i\end{array}\right)}\left(\begin{array}{l}15\\ K-i\end{array}\right)+{\displaystyle \sum _{i=0}^{k+1}\left(\begin{array}{l}12\\ i\end{array}\right)\left(\begin{array}{l}13\\ k+1-i\end{array}\right)}$  

= Equating the coefficient of $x^{k}in{\left(1+x\right)}^{25}{=}^{25}{C}_k$ .(i)

= Equating the coefficient of   $x^{k+1}in{\left(1+x\right)}^{12}{\left(x+1\right)}^{13}$

From (i) and (ii)

Hence   $26\ge k+1\Rightarrow k\le 25$

But maximum value of k is not defined bonus

${\left(2x^r+\frac{1}{x^2}\right)}^{10}$

Let the constant term is (k + 1)^th term.

${\therefore }^{10}{C}_k{\left(2x^r\right)}^{10-k}.x^{-2k}$              

${=}^{10}{C}_k{2}^{10-k}.x^{10r-rk-2k}$              

              For constant term 10r – rk – 2k = 0.

$\therefore k=\frac{10r}{r+2}k\in I$              

$\therefore r=3,k=6andr=8,k=8$               

For k = 6

For k = 8  

$\therefore r=8$

$f\left(x\right)=\frac{5x+3}{6x-\alpha }=y.....................(i)$

$\Rightarrow x=\frac{\alpha y+3}{6y-5}\Rightarrow f^{-1}\left(x\right)=\frac{\alpha x+3}{6x-5}..............(ii)$

According to question, $f\left(x\right)=f^{-1}\left(x\right)\therefore$  from (i) and (ii) we get $\propto =5$

$\frac{x+1}{a}=\frac{y}{1}=\frac{z-1}{a}...........\left(i\right)$

$\frac{x+2}{a}=\frac{y}{1}=\frac{z}{\frac{3}{b}}..........\left(ii\right)$

Let A (-1, 0, 1) and B (-2, 0, 0) $$ $\therefore$ direction ratios of AB = -1, 0, -1

$$ $\therefore$  lines are coplanar

$\therefore \left|\begin{array}{ccc}\hfill a\hfill & \hfill 1\hfill & \hfill a\hfill \\ \hfill 3\hfill & \hfill 1\hfill & \hfill \frac{3}{b}\hfill \\ \hfill -1\hfill & \hfill 0\hfill & \hfill -1\hfill \end{array}\right|=0\Rightarrow b=1anda\in R-\left\{0\right\}$

2ⁿ = 4096 -> n = 12

¹²C₆ = 924 (the greatest coefficient)

$a^{10}{b}^{10}{\left(\frac{1}{b}+\frac{2}{a}+4\right)}^{10}$

$a^{10}{b}^{10}\frac{10!{\left(\frac{1}{b}\right)}^{r_1}{\left(\frac{2}{a}\right)}^{r_2}.4^{10-r_1-r_2}}{r_1!r_2!\left(10-r_1-r_2\right)!}$

r₁ = 2, r₂ = 3

$a^7{b}^{8}is\frac{10!2^3.4^{10-2-3}}{2!3!5!}=\frac{2^{13}.10!}{2!3!5!}=2^{16}.315$    

k = 315

$\frac{{\left(2i\right)}^n}{{\left(1-i\right)}^{n-2}}$

$=\frac{{\left(2i\right)}^n{\left(1+i\right)}^{n-2}}{{\left(1+1\right)}^{n-2}}$            

$=-4{\left(-1+i\right)}^{n-2}$              

  $now{\left(-1+i\right)}^2=-2i$

$\therefore -4{\left(-1+i\right)}^{n-2}\in \left(+\right)ve$  integer for n – 2 = 4

->n = 6

 = coefficient of x⁴ in (1 + x)²¹ + coefficient of x⁴ in (1 + x)²¹

${=}^{21}{C}_4{+}^{21}{C}_4=2{.}^{21}{C}_4$              

$A_3$ = coefficient of x³ in (1 + x)²¹ + coefficient of x³ in (1 + x)²¹

${=}^{10}{C}_r.2^{10-r}.3^r.x^{30-3r-r}$  

  $For{x}^{30-4r}\to x^{\left(even\right)}$              

r = 0, 1, 2, 3, 4, 5, 6, 7

$r\in \left\{0,1,2,......,10\right\}$          

Sum of all such coefficient