Question

If n is natural number then ${{9}^{2n}}-{{4}^{2n}}$is always divisible by:
A. 13
B. both 5 and 13
C. 5
D. none of the above

Answer 1

Hint: Factorise the given expression ${{9}^{2n}}-{{4}^{2n}}$ using the formula ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$, then use the formula ${{a}^{n}}-{{b}^{n}}=\left( a-b \right)\left( {{a}^{n-1}}+{{a}^{n-2}}b+......+{{b}^{n-2}}a+{{b}^{n-1}} \right)$ to further expand the factor.

Complete step-by-step answer:
We are given the term ${{9}^{2n}}-{{4}^{2n}}$ . We know, we can write ${{9}^{2n}}-{{4}^{2n}}$ as ${{\left( {{9}^{n}} \right)}^{2}}-{{\left( {{4}^{n}} \right)}^{2}}$ , which is of the form ${{a}^{2}}-{{b}^{2}}$ . We know the identity ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ . So, we can write ${{\left( {{9}^{n}} \right)}^{2}}-{{\left( {{4}^{n}} \right)}^{2}}$ as $\left( {{9}^{n}}+{{4}^{n}} \right)\left( {{9}^{n}}-{{4}^{n}} \right)$ .
Now, we can see that there are two factors of ${{9}^{2n}}-{{4}^{2n}}$ , i.e. $\left( {{9}^{n}}+{{4}^{n}} \right)$ and $\left( {{9}^{n}}-{{4}^{n}} \right)$ . Let’s see the first factor, i.e. $\left( {{9}^{n}}+{{4}^{n}} \right)$ . We know that we can expand $\left( {{9}^{n}}+{{4}^{n}} \right)$ using the formula ${{a}^{n}}+{{b}^{n}}=\left( a+b \right)\left( {{a}^{n-1}}-{{a}^{n-2}}b+......-{{b}^{n-2}}a+{{b}^{n-1}} \right)$ , when n is odd natural number. But we are given that n is any natural number. So, we cannot expand $\left( {{9}^{n}}+{{4}^{n}} \right)$. So, let’s see the second factor, i.e. $\left( {{9}^{n}}-{{4}^{n}} \right)$ . We know, we can expand $\left( {{9}^{n}}-{{4}^{n}} \right)$ using the formula ${{a}^{n}}-{{b}^{n}}=\left( a-b \right)\left( {{a}^{n-1}}+{{a}^{n-2}}b+......+{{b}^{n-2}}a+{{b}^{n-1}} \right)$ .
$\therefore \left( {{9}^{n}}-{{4}^{n}} \right)=\left( 9-4 \right)\left( {{9}^{n-1}}+{{9}^{n-2}}.4+{{.....4}^{n-2}}a+{{4}^{n-1}} \right)$
$\therefore {{9}^{2n}}-{{4}^{2n}}=\left( {{9}^{n}}+{{4}^{n}} \right)\left( 5 \right)\left( {{9}^{n-1}}+{{9}^{n-2}}.4+{{......9.4}^{n-2}}+{{4}^{n-1}} \right)$
Here, we can clearly see that 5 comes in the factor of ${{9}^{2n}}-{{4}^{2n}}$. So, we can say that ${{9}^{2n}}-{{4}^{2n}}$ is divisible by 5.
Therefore, we get the answer as option (C).

Note:While factorizing, make sure to use correct identities. Also, while doing calculations, take care of the signs, as sign mistakes are very common and can lead to wrong answers. The identities used in the problem must be remembered perfectly as confusion in the identities can lead to wrong answers. Students generally forget that ${{a}^{n}}+{{b}^{n}}=\left( a+b \right)\left( {{a}^{n-1}}-{{a}^{n-2}}b+......-{{b}^{n-2}}a+{{b}^{n-1}} \right)$ is valid only when n is odd number and mark option B as the answer, which is wrong. Such confusions should be avoided.