Maths Binomial Theorem

${\alpha }_n=19^n-12^n$

$\frac{31{\alpha }_9-{\alpha }_{10}}{57.{\alpha }_2}=\frac{31\left(19^9-12^9\right)-\left(19^{10}-12^{10}\right)}{57\left(19^8-12^8\right)}$

$=\frac{19^9\left(31-19\right)-12^9\left(31-12\right)}{57\left(19^8-12^8\right)}$

$=\frac{19^9.12-12^9.19}{57\left(19^8-12^8\right)}=4$

${\displaystyle \sum _{k=1}^{10}\left(2k-1\right).k{.}^{10}{C}_k}$

$=2\sum {k^2}^{10}{C}_k-\sum k{.}^{10}{C}_k$

$x=1\Rightarrow {\displaystyle \sum _{k=0}^{10}k}{.}^{10}{C}_k=10.2^9$

$x=1\Rightarrow \sum k^2{}^{10}{C}_k=90.2^8+10.2^9=2^8\left(90+201\right)=2^8.110$

S = 2⁹ . 110 – 10.2⁹ = 2⁹ . 100

S = 2⁹ . 100

$\frac{2^{11}.25}{2^1-1}$

$t_n=\frac{1}{2^n.3^{11-n}}={\left(\frac{1}{3}\right)}^{11}.{\left(\frac{3}{2}\right)}^n$

$S_{10}={\left(\frac{1}{3}\right)}^{11}.\frac{\left(\frac{3}{2}\right).\left({\left(\frac{3}{2}\right)}^{10}-1\right)}{\frac{3}{2}-1}$

$=\frac{1}{3^{10}}.\frac{3^{10}-2^{10}}{2^{10}}$

$=1{+}^{10}{C}_1.2^1{+}^{10}{C}_2.2^2+....{+}^{10}{C}_9.2^9$

k = 21 + 6m ® R = 3

= A + B (say)

(1 + x)¹⁵ =

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

put  = x = -1

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

  $B{=}^{14}{C}_{13}=2^{13}$            

$\therefore A+B=2^{13}-14$           

2²? ¹ - 1 - 1 = 126
2²? ¹ = 2?
2n + 1 = 7
n = 3

(1 + x)? = C? + C? x + C? x² + . C? x?
Put x = 1
2? = C? + C? + C? + C? + … C?
Put x = ω
(1 + ω)? = C? + C? ω + C? ω² + . C? (ω)?
Put x = ω²
(1 + ω²)? = C? + C? ω² + C? ω² + … C? (ω²)?
∴ 1 + ω + ω² = 0
Now add all three equation (s)
2? + (1 + ω)? + (1 + ω²)? = 3 [C? + C? + C? + …]
C? + C? + C? + C? + …
= (2? + (1+ω)? + (1+ω²)? ) / 3
= (2? + (1+ω)? + (-ω)? ) / 3 = (2? - (-1)? ) / 3

[3¹/²]^ (log? (25? ¹+7)+ [3¹/? ]^ (log? (5? ¹+1) = 180
⇒ 45 (5²? ²+7) / (5? ¹+1) = 180
⇒ (5²? ²+7)/ (5? ¹+1) = 4
Put 5? ¹=t
⇒ (t²+7)/ (t+1) = 4
⇒ t²−4t+3=0
⇒t=1,3
⇒5? ¹=1 or 5? ¹=3
⇒x=1 or x−1=log?3

T? =? C? (x¹/³)? (x? ¹/? )? (5¹/²)? (5? ¹)?
for x¹? : (60-r)/3 - r/5 = 10
=> 180 – 3r – 2r = 60
=> r = 24
k = 3 + exponent of 5 in? C?
= 3 + [60/5] + [60/25] – [24/5] – [24/25] – [36/5] – [36/25]
= 3 + (12+2–4–0–7–1)
= 3+2=5

cos? ¹ (y/2) = log? (x/5)? , |y| < 2
Differentiating on both side
(1/√ (1- (y/2)²) * (-y'/2) = 5 * (5/x) * (1/5)
(-xy')/ (2√ (1- (y²/4) = 5
Square on both side
(x²y'²)/4 = 25 * (4-y²)/4
Diff on both side
xy' + y'x² + 25y = 0