(a + b)² can be expanded using binomial orem to get:
(a + b)² = a² + 2ab + b²
Here, a = xy and b = 2
Therefore, (xy + 2)² = (xy)² + 2(xy)(2) + (2)²
(xy + 2)² = x²y² + 4xy + 4.

Clearly (2n + 1) is an odd number. Therefore this is a case of binomial with an odd power.
For a binomial expansion with odd power, re are two middle terms.
Case 1: \(\frac{n+1}{2}\) = 19
Therefore n = 37
Case 2: \(\frac{n+3}{2}\) = 19
Therefore n = 35

The numerator when simplified is of form (101 ? 99)³
The denominator can be simplified as (101 ? 99)(101 + 99)
When we substitute in numerator and denominator we get (2 x 2 x 2) / (2 x 200)
= 1/50.

The coefficient of first term and last term is same. The first term is nC₁ aⁿ and last term is nC₀ bⁿ unless, a and b are numbers that change value of coefficient.

Using binomial orem (9 + 3i)² = 81 + 54i + 9i²
We know that i² = ?1
Therefore, (9 + 3i)² = 81 + 54i ? 9
(9 + 3i)² = 72 + 54i
Real part = 72.

4¹⁰³ = 4 x 4¹⁰²
4¹⁰³ = 4 x (4²)⁵¹
4¹⁰³ = 4 x (16)⁵¹
4¹⁰³ = 4 x (17 ? 1)⁵¹
4¹⁰³ = 4 x \(\Sigma_{r = 0}^{r = 51}\)(51Cr 17^{24 ? r} (-1)^r
4¹⁰³ = 4 x [51C₀ 17⁵¹ (-1)⁰ + 51C₁ 17⁵¹ (-1)¹ +?.+ 51C₅₀ 17¹ (-1)⁵⁰ + 51C₅₁ 17⁰ (-1)⁵¹]
4¹⁰³ = 4 x 17 x k ? 1
The remainder = 17 ? 1
Remainder = 16.

Since power is even, re are odd number of terms.
The middle term is \((\frac{n}{2} + 1)^{th}\) term.
= \((\frac{80}{2} + 1)^{th}\) term
= 41^st term
The 41^st term = ⁸⁰C₄₀ (xyz)⁴⁰ (3)⁴⁰

x² + y² + 2xy is expansion of (x + y)²
x² ? y² can be written as (x ? y)(x + y)
Substituting in fraction we get, \(\frac{(x + y)^2}{(x ? y)(x + y)}\).
After cancelling terms we get, \(\frac{x^2+y^2+2xy}{x^2-y^2}\) = (x + y) (x ? y)^-1.

If n is number of terms and is even, n middle term is \((\frac{n}{2} + 1)^{th}\) term.
Else if n is number of terms and is odd, re are two middle terms which are \((\frac{n + 1}{2})^{th}\) term and \((\frac{n + 3}{2})^{th}\) term.
Case 1: \(\frac{n}{2} + 1\) = 5 ; n = 8
Case 2: \(\frac{n + 1}{2}\) = 5 ; n = 9
Case 3: \(\frac{n + 3}{2}\) = 5 ; n = 7

Since middle term is fourth term
Considering 3x to be even, \((\frac{3x + 1}{2})^{th}\) term = 4
x = 7/3
Therefore, fourth term coefficients are ⁷C₃