JAMIA 2023 — Mathematics PYQ

We can rewrite the first term by taking a common denominator:

Now, substitute this back into the original expression:

So, the expression is equivalent to:

Step 2: Identify the Middle Term

The middle term of the original expression depends on the middle term of the expansion of $(1+x^2{)}^{n+1}$.

Case 1: If $(n+1)$ is even (i.e., $n$ is odd)

There is one middle term at position $r=\frac{n+1}{2}$.

The term in the expansion is:

Dividing by the $x^2$ outside the bracket:

Case 2: If $(n+1)$ is odd (i.e., $n$ is even)

There are two middle terms at positions $r=\frac{(n+1)-1}{2}=\frac{n}{2}$ and $r=\frac{(n+1)+1}{2}=\frac{n}{2}+1$.

1. First Middle Term ($r=\frac{n}{2}$):

   $$\frac{\left(\genfrac{}{}{0ex}{}{n+1}{n/2}\right)(x^2{)}^{n/2}}{x^2}=\left(\genfrac{}{}{0ex}{}{n+1}{n/2}\right)x^{n-2}$$

2. Second Middle Term ($r=\frac{n}{2}+1$):

   $$\frac{\left(\genfrac{}{}{0ex}{}{n+1}{n/2+1}\right)(x^2{)}^{n/2+1}}{x^2}=\left(\genfrac{}{}{0ex}{}{n+1}{n/2+1}\right)x^n$$

General Form (Often found in options)

Often, this question is solved by looking for the term independent of $x$ or a specific coefficient. If the question implies a single general coefficient form:

The middle term of $(1+x^2{)}^{n+1}$ is $\left(\genfrac{}{}{0ex}{}{n+1}{k}\right)x^{2k}$. When divided by $x^2$, the coefficient remains $\left(\genfrac{}{}{0ex}{}{n+1}{k}\right)$.

Final Answer:

If $n$ is odd, the middle term is:

If $n$ is even, the middle terms are: