Q.Select the correct answer from the given alternatives. The middle term in the expansion of $(1+x)^{2n}$ will be: (A) $(n-1)$th (B) $n$th (C) $(n+1)$th (D) $(n+2)$th

Because the expansion of $(a+b)^n$ always has $n+1$ terms, whether that count is odd or even decides how many terms sit exactly in the middle. If $n$ is even, $n+1$ is odd, so there is a single, uniquely-determined middle term: the $\left(\dfrac n2+1\right)$th term. If $n$ is odd, $n+1$ is even, so there is no single middle term — instead there are two middle terms, the $\left(\dfrac{n+1}2\right)$th and the $\left(\dfrac{n+3}2\right)$th, which are consecutive terms straddling the centre of the expansion. Once this position (or pair of positions) has been identified purely from the parity of $n$, the value of the middle term is found in exactly the same way as any other term — by substituting the matching value of $r$ into the general term formula $t_{r+1}={}^nC_ra^{n-r}b^r$. So this idea is really a two-step recipe layered on top of the general term: first a quick parity check to find which term(s) to compute, then the ordinary general-term substitution to compute it. It shows up throughout the chapter's exercises on expressions such as $\left(x^2+\dfrac1x\right)^7$ (odd exponent, two middle terms) and $\left(x^2-\dfrac2x\right)^8$ (even exponent, one middle term), and in reverse problems where a stated numeric value of the middle term (e.g. 'the middle term is 160') is used to solve for an unknown constant inside the binomial.