Maths Binomial Theorem

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

${\left(\frac{x}{4}-\frac{12}{x^2}\right)}^{12}$

According to question, 12 – 3r = 0 r = 4

$\Rightarrow \frac{3^6}{4^4}*55=\frac{3^6}{4^4}*k\Rightarrow k=55$

Kindly consider the following figure

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

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

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

  $now{\left(-1+i\right)}^2=-2i$

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

  $S=\frac{{\left(5+x\right)}^{500}.\left(1-{\left(\frac{x}{5+x}\right)}^{501}\right)}{1-\frac{x}{5+x}}$  

  $=\frac{{\left(5+x\right)}^{500}-\frac{x^{501}}{5+x}}{\frac{5}{\left(5+x\right)}}=\frac{{\left(5+x\right)}^{501}-x^{501}}{5}$              

Coefficient of x¹⁰¹  

$tn=\frac{2\left(2^3+4^3+6^3+....+{\left(2n\right)}^3\right)-\left(1^3-2^3+3^3+.....{\left(2n\right)}^3\right)}{n\left(4n+3\right)}$

^{$\frac{2*2^3*{\left(\frac{n*\left(n+1\right)}{2}\right)}^2-{\left(\frac{2n*\left(2n+1\right)}{2}\right)}^2}{n\left(4n+3\right)}$}

^{tn = n}

^{$now{\displaystyle \sum _{n=1}^{15}{T}_n}=\frac{15.16}{2}$}

Kindly consider the following figure