7.2 Q-7 | Mathematical Induction And Binomial Theorem | Mathematics 11

Exercise 7.2 — Question 7

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

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

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

The total number of terms in $(a + b)^{n}$ is $n + 1$ .

Finding the middle term:

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 $n = 10$ (even).

Total terms $= 10 + 1 = 11$

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

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

Finding $T_{6}$ (set $r = 5$ ):

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

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

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

$= 252$

The middle term is $252$ .

$n$	Even/Odd	Number of Middle Terms	Middle Term(s)
10	Even	1	$T_{6} = 252$