Question

please do this sum​

Answer 1

Solution :

Now, $\displaystyle\frac{{a}^{7 + 2n} \times {{a}^{(2)}}^{3n + 2} }{{{a}^{(4)}}^{2n + 3} }$

$\displaystyle=\frac{{a}^{7 + 2n} \times {a}^{6n + 4}}{ {a}^{8n + 12} }$

$\displaystyle=\frac{ {a}^{7 + 2n + 6n + 4}}{ {a}^{8n + 12} }$

$\displaystyle=\frac{{a}^{11 + 8n}}{ {a}^{8n + 12} }$

$\displaystyle={a}^{11 + 8n - 8n - 12}$

$\displaystyle={a}^{ - 1}$

$\displaystyle=\frac{1}{a}$

$\displaystyle\implies \boxed{\frac{{a}^{7 + 2n} \times {{a}^{(2)}}^{3n + 2} }{{{a}^{(4)}}^{2n + 3}}=\frac{1}{a}}$

STEPS FOR UNDERSTANDING :-

1. First of all we shall open the powers which are over exponents by multiplying them. like :- (a^b)^c = a^bc both in numerator and denominator.

2. Next we shall multiply the numbers having same base "a" in the numerator by adding their exponents.

3. We shall now divide the fractions having same base by subtracting the exponents.

4.For these numericals please follow below mentioned rules.

Rules used above :

$\displaystyle{a}^{ - b} = \frac{1}{ {a}^{b} }$

$\displaystyle{a}^{b} \times {a}^{c} = {a}^{b + c}$

$\displaystyle{ {a}^{(b)}}^{c} = {a}^{bc}$

$\displaystyle\frac{ {a}^{b} }{ {a}^{c} } = {a}^{b - c}$