Maths Binomial Theorem

Kindly go through the solution

BONUS

Ratio = ²? C? / (¹? C? + ¹? C? ) = ²? C? / ²? C? = 1.

(x+1)/ (x²/³-x¹/³+1) - (x-1)/ (x-x¹/²) = (x¹/³+1) - (1+x? ¹/²)
= (x¹/³ - x? ¹/²)¹?
general term = ¹? C? (x¹/³)¹? (-x? ¹/²)? = ¹? C? x^ (20-5r)/6)
For independent of x, 20-5r=0 ⇒ r=4
∴ Coefficient = ¹? C? = 210

Σ (1/a) (1-rb/a)? ¹ = (1/a)Σ (1+rb/a+r²b²/a²+.)
≈ (1/a)Σ (1+rb/a) = n/a + (b/a²)n (n+1)/2
Compare coeffs: α=1/a, β=b/2a². γ=b²/3a³. This differs from solution.

We know, ?C? is max at middle term
a = ¹?C? = ¹?C? = ¹?C?
b = ²?C_q = ²?C?
c = ²¹C? = ²¹C? = ²¹C?
a/¹?C? = b/(²?C?) = c/(²¹C?) = 20/10, 21/11

(1 + x)¹? + x(1+x)? + x²(1+x)? + . . + x¹?
= (1-x)¹? [1-(x/(1+x))¹¹]/[1-x/(1+x)]
⇒ (1+x)¹¹ - x¹¹
Coefficient of x? is ¹¹C? = 330

120.00
(1 + x + x² + x?)? = (1 + x)? · (1 + x²)?
Coefficient of x? = ?C?·?C? + ?C?·?C? + ?C?·?C?
= 15 + 90 + 15
= 120

Sum S = Σ (2r+1)? C? = 2Σr? C? + Σ? C? = 2n2? ¹ + 2? = n2? + 2? = (n+1)2?
(n+1)2? = 101 * 2¹? ⇒ n=100.
2 [ (n-1)/2] = 2 [ (99)/2] = 2 [49.5] = 2*49 = 98.

Coeff of x? in (2+x/3)? is? C? 2? (1/3)?
Coeff of x? in (2+x/3)? is? C? 2? (1/3)?
? C? 2? / 3? =? C? 2? / 3?
(n!/ (7! (n-7)!) * 2 = (n!/ (8! (n-8)!) * (1/3).
2 / (n-7) = 1 / (8*3).
48 = n-7 ⇒ n=55.

T? = ¹? C? (xsinα)¹? (acosα/x)? = ¹? C? x¹? ²? sin¹? α a? cos? α
For term independent of x, 10-2r=0 ⇒ r=5.
T? = ¹? C? sin? α a? cos? α = ¹? C? (sin2α/2)? a?
This is the greatest when sin2α=1.
the greatest value = ¹? C? (a/2)? = 10!/ (5!)².
¹? C? = 252.
252 (a/2)? = 252.
(a/2)? = 1 ⇒ a=2.