Maths Binomial Theorem

$9^n-8n-1={\left(1+8\right)}^n-8n-1$  

$=(1+8n{+}^n\text{?}{C}_2\cdot 8^2{+}^n\text{?}{C}_3\cdot 8^3+....]-8n-1$  

So, a = ⁿC₂ + ⁿC₃8 + ⁿC₄8² +….

Similarly, b = ⁿC₂ + ⁿC₃ 5 + ⁿC₄ 5² +….

a - b = ⁿC₃ (8 – 5) + ⁿC₄ (8² – 5²) +….

${\left(2x^3+\frac{3}{x^k}\right)}^{12}$

gen term = 

${=}^{12}{C}_r{2}^{12-r}.3^r.x^{36-3r-rk}$  

For constant term

36 – 3r – rk = 0

$\Rightarrow k=\frac{36-3r}{r}$  

for r = 1, 2, 4   

Possible values of k = 3, 1

$T_{r+1}{=}^{15}\text{?}{C}_r\cdot {\left(2x^{1/5}\right)}^{15-r}{\left(-x^{-1/5}\right)}^r$  

= ¹⁵C_{r $\cdot 2^{15-r}\cdot x^{\frac{15-2r}{5}}\cdot {\left(-1\right)}^r$}

Coefficient of x^-1 -> r = 10 ->m = ¹⁵C_{10 $\cdot 2^5$}

$x^{-3}\Rightarrow r=15\Rightarrow n{=}^{15}\text{?}{C}_{15}\cdot 2^0=1$  

now mn² = ¹⁵C_{r $\cdot 2^r$}  

r = 5

Coefficient of x in ${\left(1+x\right)}^p{\left(1-x\right)}^q={-}^p{C_0}^q{C}_1{+}^p{C_1}^q{C}_0=-3\Rightarrow p-q=3$  

              Coefficient of x² in ${\left(1+x\right)}^p{\left(1-x\right)}^q{=}^p{C_0}^q{C}_2{-}^p{C_1}^q{C}_1{-}^p{C_2}^q{C}_0=-5$  

$\Rightarrow \frac{q\left(q-1\right)}{2}-pq+\frac{p\left(p-1\right)}{2}=-5\Rightarrow \frac{q\left(q-1\right)}{2}-\left(q-3\right)q+\frac{\left(q-3\right)\left(q-4\right)}{2}=-5\Rightarrow q=11,p=8$  

              Coefficient of x³ in ${\left(1+x\right)}^8{\left(1-x\right)}^{11}={-}^{11}{C}_3{+}^8{C_1}^{11}{C}_2{-}^8{C_2}^{11}{C}_1{+}^8{C}_3=23$  

$T_{r+1}{=}^{15}{C}_r{\left(x^2\right)}^{15-r}{\left(\frac{3\lambda }{x}\right)}^r$

$T_{r+1}{=}^{15}{C}_4{\left(3\lambda \right)}^r.x^{30-3r}$

30 – 3r = 21 =>$$ r = 3

${\lambda }^3=\frac{8}{27}\Rightarrow \lambda =\frac{2}{3}$

$1+2x+3x^2+4x^3\dots =(1-x{)}^{-2}$

${\left(1+2x+3x^2+4x^3\dots \right)}^{-2}=(1-x{)}^4=1-4x+6x^2-4x^3+x^4$

$a_0=1,\mathrm{}{a}_1=-4$

 $a_0-a_1=5$

  $a_0{C}_0-a_1{C}_1+a_2{C}_2-a_3{C}_3+.....+a_{2012}{c}_{2012}$

$=a_0\left(C_0-C_1+C_2+....+C_{2012}\right)$            

  $-d\left(C_1-2C_2+3C_3-4C_4+....{-}^{2012}{C}_{2012}\right)$           

  $=a_0\left(0\right)-d\left(C_1-2C_2+3C_3{-}^4{C}_4\right)$           

+ …….-  

 =   $a_0\left(0\right)-d\left(0\right)$

= 0

$(x+y{)}^n={}^n{C}_r{x}^{n-r}{y}^r$

${\left(\frac{3}{2}{x}^2+\frac{2}{x}\right)}^{12}?{}^{12}{C}_r{\left(\frac{3}{2}{x}^2\right)}^{12-r}{\left(\frac{2}{x}\right)}^r$

For independent of X .

$x^{24-2r}{x}^{-r}=x^0;24=3r\Rightarrow r=8$

${}^{12}{C}_8{\left(\frac{3}{2}\right)}^{12-8}(2{)}^8;\frac{\lfloor 12}{\overline{12}44}*\frac{3^4}{2^4}*2^8\Rightarrow \lambda ;\frac{\lambda }{81}\Rightarrow 7920$

(5³)^740 * 5² = 5² (125)^740 = 5² (126 – 1)^740
So remainder = 4.

log x = t
(t² - 1/t)¹²
T? = ¹²C? (t²)¹²? (-1/t)? = ¹²C? t²? ³? (-1)?
For constant term r = 8 & coefficient ¹²C?
= (12*11*10*9) / 24 = 45*11 = 495