7.2 Q-2 | Mathematical Induction And Binomial Theorem | Mathematics 11

Exercise 7.2 — Question 2

This exercise applies the Binomial Theorem to find specific terms, middle terms, and simplify binomial expansions.

For the expansion of $(a + b)^{n}$ , the general term (the $(r + 1)$ -th term) 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 $n + 1$ .

$n$	Number of middle terms	Middle term(s)
Even	1	$T_{\frac{n}{2} + 1}$
Odd	2	$T_{\frac{n + 1}{2}}$ and $T_{\frac{n + 3}{2}}$

Example 1: Find the middle term of $(x + \frac{1}{x})^{6}$ .

Solution:

Here $a = x$ , $b = \frac{1}{x}$ , $n = 6$ (even)
Middle term: $T_{\frac{6}{2} + 1} = T_{4}$
$T_{4} = (3 6) x^{6 - 3} \cdot (\frac{1}{x})^{3} = 20 \cdot x^{3} \cdot \frac{1}{x ^{3}} = 20$

Example 2: Find the term independent of $x$ in $(x^{2} + \frac{1}{x})^{9}$ .

Solution:

$T_{r + 1} = (r 9) (x^{2})^{9 - r} \cdot (\frac{1}{x})^{r} = (r 9) x^{18 - 2 r} \cdot x^{- r} = (r 9) x^{18 - 3 r}$
For the term independent of $x$ : $18 - 3 r = 0 \Rightarrow r = 6$
$T_{7} = (6 9) = (3 9) = 84$

Example 3: Find the two middle terms of $(2 x - y)^{5}$ .

Solution:

$n = 5$ (odd), so two middle terms: $T_{\frac{5 + 1}{2}} = T_{3}$ and $T_{\frac{5 + 3}{2}} = T_{4}$
$T_{3} = (2 5) (2 x)^{3} (- y)^{2} = 10 \cdot 8 x^{3} \cdot y^{2} = 80 x^{3} y^{2}$
$T_{4} = (3 5) (2 x)^{2} (- y)^{3} = 10 \cdot 4 x^{2} \cdot (- y^{3}) = - 40 x^{2} y^{3}$

This exercise applies the Binomial Theorem to find specific terms, middle terms, and simplify binomial expansions.

For the expansion of $(a + b)^{n}$ , the general term (the $(r + 1)$ -th term) 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 $n + 1$ .

$n$	Number of middle terms	Middle term(s)
Even	1	$T_{\frac{n}{2} + 1}$
Odd	2	$T_{\frac{n + 1}{2}}$ and $T_{\frac{n + 3}{2}}$

Example 1: Find the middle term of $(x + \frac{1}{x})^{6}$ .

Solution:

Here $a = x$ , $b = \frac{1}{x}$ , $n = 6$ (even)
Middle term: $T_{\frac{6}{2} + 1} = T_{4}$
$T_{4} = (3 6) x^{6 - 3} \cdot (\frac{1}{x})^{3} = 20 \cdot x^{3} \cdot \frac{1}{x ^{3}} = 20$

Example 2: Find the term independent of $x$ in $(x^{2} + \frac{1}{x})^{9}$ .

Solution:

$T_{r + 1} = (r 9) (x^{2})^{9 - r} \cdot (\frac{1}{x})^{r} = (r 9) x^{18 - 2 r} \cdot x^{- r} = (r 9) x^{18 - 3 r}$
For the term independent of $x$ : $18 - 3 r = 0 \Rightarrow r = 6$
$T_{7} = (6 9) = (3 9) = 84$

Example 3: Find the two middle terms of $(2 x - y)^{5}$ .

Solution:

$n = 5$ (odd), so two middle terms: $T_{\frac{5 + 1}{2}} = T_{3}$ and $T_{\frac{5 + 3}{2}} = T_{4}$
$T_{3} = (2 5) (2 x)^{3} (- y)^{2} = 10 \cdot 8 x^{3} \cdot y^{2} = 80 x^{3} y^{2}$
$T_{4} = (3 5) (2 x)^{2} (- y)^{3} = 10 \cdot 4 x^{2} \cdot (- y^{3}) = - 40 x^{2} y^{3}$