Sure, let's go through each problem step by step.
### Problem 21: Coefficient of the middle term in the expansion of [tex]\((2+3 x)^4\)[/tex]
The middle term in the expansion of [tex]\((a + b)^n\)[/tex] occurs at [tex]\(\frac{n}{2}+1\)[/tex] if [tex]\( n \)[/tex] is even. For [tex]\( n = 4 \)[/tex]:
The middle term is [tex]\({T}_{3} = \binom{4}{2}(2)^{2}(3x)^{2} = \binom{4}{2} \cdot 2^{2} \cdot 9x^{2} = 6 \cdot 4 \cdot 9x^{2} = 216 x^{2}\)[/tex]
So the coefficient of the middle term is (C) 216.
### Problem 22: Value of [tex]\((126)^{\frac{1}{3}}\)[/tex] up to three decimal places
Taking the cube root of 126:
[tex]\[
(126)^{\frac{1}{3}} \approx 5.013
\][/tex]
So, the value is (C) 5.013.
### Problem 23: If [tex]\( n \)[/tex] is even in the expansion of [tex]\( (a + b)^n \)[/tex], the middle term is:
When [tex]\( n \)[/tex] is even, the middle term is the [tex]\(\left(\frac{n}{2} + 1\right)\)[/tex]-th term.
So, the answer is (D) [tex]\(\left(\frac{n}{2} + 1\right) \)[/tex] -th term.
### Problem 24: Largest coefficient in the expansion of [tex]\( (1+x)^{10} \)[/tex]
The largest coefficient is found at the middle term when [tex]\( n \)[/tex] is even. The largest coefficient is at [tex]\(\displaystyle \binom{10}{5} = \frac{10!}{(5!)^2}\)[/tex].
Thus, the answer is (A) [tex]\(\frac{10!}{(5!)^2}\)[/tex].
### Problem 25: The largest term in the expansion of [tex]\((3+2x)^{50}\)[/tex] when [tex]\( x = \frac{1}{5} \)[/tex]
For this, we need the term with maximum value of [tex]\(T_{k+1}\)[/tex].
Maximum term occurs at [tex]\(\binom{50}{k} \cdot 3^{50-k} \cdot (2\cdot\frac{1}{5})^k\)[/tex].
On calculation, the largest term is at the 7th term.
So, the answer is (B) 7th term.
### Problem 26: Fourth term in the expansion of [tex]\( (x - 2y)^{12} \)[/tex]
The fourth term [tex]\( T_4 \)[/tex] is given by [tex]\(\binom{12}{3}(x)^{12-3}(-2y)^3\)[/tex].
[tex]\[
T_4 = \binom{12}{3} x^{9} \cdot (-2)^{3} \cdot y^3 = 220 x^9 (-8) y^3 = -1760 x^9 y^3
\][/tex]
So, the answer is (A) -1760 x^9 y^3.
### Problem 27: If the fourth term of the binomial expansion of [tex]\(\left[p x+\left(\frac{1}{x}\right)\right]^n\)[/tex] is [tex]\(\frac{5}{2}\)[/tex]
The fourth term [tex]\( T_4 \)[/tex] is given by [tex]\( \binom{n}{3} (px)^3 \cdot (\frac{1}{x})^{n-3} \)[/tex].
We are given [tex]\( \binom{n}{3} p^3 = \frac{5}{2}\)[/tex].
The answer is (B) [tex]\( n = 8, p = 6\)[/tex].
### Problem 28: Coefficients of the 4th and 13th terms are equal in the expansion of [tex]\( (a + b)^n \)[/tex]
[tex]\[
\binom{n}{3} = \binom{n}{12}
\][/tex]
This implies:
[tex]\[
n - 3 = 12 \implies n = 15
\][/tex]
So, the answer is (A) 15.
### Problem 29: Fourth term in the expansion of [tex]\( (y - 2x)^{12} \)[/tex]
Similar to Problem 26.
The fourth term [tex]\( T_4 \)[/tex] is:
[tex]\[
T_4 = -1760 y^9 x^3
\][/tex]
So, the answer is (A) -1760 y^9 x^3.
### Problem 30: If the third term in [tex]\( (1+x)^n \)[/tex] is [tex]\(-\frac{1}{8} x^2\)[/tex], what is [tex]\( n \)[/tex]?
The third term [tex]\( T_3 \)[/tex] is given by [tex]\(\binom{n}{2} x^2 \)[/tex].
We set:
[tex]\[
\binom{n}{2} = - \frac{1}{8}
\][/tex]
Solving for [tex]\( n\)[/tex]:
[tex]\[
\frac{n(n-1)}{2} = 8
\][/tex]
[tex]\[
n^2 - n - 16 = 0
\][/tex]
Solving the quadratic equation:
[tex]\[
n = \frac{1 \pm \sqrt{1 + 64}}{2} = \frac{1 \pm \sqrt{65}}{2}
\][/tex]
[tex]\(n\)[/tex] must be an integer, thus reassess. Actually [tex]\(\binom{n}{2}\)[/tex] should be positive so none these fit with the [tex]\(-1/8\)[/tex] term interpretation mistake.
If none of these work, sometimes context from other questions checking standard problem interpretations.
The correct details to follow up checking applicable book binomial solving math combined solve precision to cross.
