7.2 Q-5 | Mathematical Induction And Binomial Theorem | Mathematics 11

Exercise 7.2 — Question 5

For the binomial expansion of $(a + b)^{n}$ , the total number of terms is $n + 1$ .

Rule for Middle Terms:

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 $T_{\frac{n + 3}{2}}$

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

Here $n = 6$ (even), $a = x$ , $b = \frac{1}{x}$ .

Middle term = $T_{\frac{6}{2} + 1} = T_{4}$

$T_{4} = T_{3 + 1} = (3 6) x^{6 - 3} (\frac{1}{x})^{3} = 20 \cdot x^{3} \cdot \frac{1}{x ^{3}} = 20$

Here $n = 7$ (odd), $a = x$ , $b = - \frac{1}{2 x}$ .

Two middle terms: $T_{\frac{7 + 1}{2}} = T_{4}$ and $T_{\frac{7 + 3}{2}} = T_{5}$

$T_{4} = (3 7) x^{4} (- \frac{1}{2 x})^{3} = 35 \cdot x^{4} \cdot (- \frac{1}{8 x ^{3}}) = - \frac{35 x}{8}$

$T_{5} = (4 7) x^{3} (- \frac{1}{2 x})^{4} = 35 \cdot x^{3} \cdot \frac{1}{16 x ^{4}} = \frac{35}{16 x}$