Binomial Theorem

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.

Coeff of x? in (x²+1/bx)¹¹: T? = ¹¹C? (x²)¹¹? (1/bx)? = ¹¹C? x²²? ³? b?
22-3r=7 ⇒ 3r=15 ⇒ r=5. Coeff is ¹¹C? /b?
Coeff of x? in (x-1/bx²)¹¹: T? = ¹¹C? (x)¹¹? (-1/bx²)? = ¹¹C? x¹¹? ³? (-1)? b?
11-3r=-7 ⇒ 3r=18 ⇒ r=6. Coeff is ¹¹C? (-1)? /b? = ¹¹C? /b?
¹¹C? /b? = ¹¹C? /b? ⇒ b = ¹¹C? /¹¹C? = (11-5+1)/5 = 7/5. This differs from the solution.
Let's check the exponents again.
x? : 22-2r-r=7 => 22-3r=7 => 3r=15 => r=5. Correct.
x? : 11-r-2r=-7 => 11-3r=-7 => 3r=18 => r=6. Correct.
The given solution has b=1.

$T_{r+1}{=}^{10}{C}_r{\left(t{x}^{\frac{1}{5}}\right)}^{10-r}{\left(\frac{{\left(1-x\right)}^{\frac{1}{10}}}{t}\right)}^r$

According to question, 10 – 2r = 0 ⇒ r = 5

$\therefore T_6{=}^{10}{C}_5*{\left(1-x\right)}^{\frac{1}{2}}$

T₆ is maximum, when f(x) = $x{\left(1-x\right)}^{\frac{1}{2}}$ is maximum.

$f\text{'}\left(x\right)={\left(1-x\right)}^{\frac{1}{2}}-\frac{x}{2\sqrt[]{1-x}}=\frac{2\left(1-x\right)-x}{2\sqrt[]{1-x}}$

For maximum, $f\text{'}\left(x\right)=0\Rightarrow x=\frac{2}{3}$

$\therefore T_6{=}^{10}{C}_5\frac{2}{3}.{\left(\frac{1}{3}\right)}^{\frac{1}{2}}=\frac{2\left(10!\right)}{3\sqrt[]{3}{\left(5!\right)}^2}$