7.2 Q-13 | Mathematical Induction And Binomial Theorem | Mathematics 11

Exercise 7.2 — Question 13

Find the middle term(s) in the expansion of $(x + \frac{1}{x})^{10}$ .

For the binomial expansion $(a + b)^{n}$ , the general term is:

$T_{r + 1} = (r n) a^{n - r} b^{r}, r = 0, 1, 2, \dots, n$

Finding the middle term:

Total number of terms in $(a + b)^{n}$ is $n + 1$ .
If $n$ is even, there is one middle term: $T_{\frac{n}{2} + 1}$ .
If $n$ is odd, there are two middle terms: $T_{\frac{n + 1}{2}}$ and $T_{\frac{n + 3}{2}}$ .

Here $a = x$ , $b = \frac{1}{x}$ , and $n = 10$ .

Since $n = 10$ is even, there is exactly one middle term:

$T_{\frac{10}{2} + 1} = T_{6}$

Using the general term formula with $r = 5$ :

$T_{6} = (5 10) x^{10 - 5} (\frac{1}{x})^{5}$

$T_{6} = (5 10) x^{5} \cdot \frac{1}{x ^{5}}$

$T_{6} = (5 10) \cdot x^{0}$

$T_{6} = 252$

The middle term is $252$ .

$n$	Number of Terms	Middle Term(s)
Even	$n + 1$ (odd)	$T_{\frac{n}{2} + 1}$
Odd	$n + 1$ (even)	$T_{\frac{n + 1}{2}}$ and $T_{\frac{n + 3}{2}}$