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Q.
If n is a +ve integer, then $ 2.4^{2n+1} + 3^{3n+1}$is divisible by
Principle of Mathematical Induction
Solution:
Let $P(n) = 2 . 4^{2n + 1} + 3^{3n + 1}$
Then $P(1) = 2 . 4^3 + 3^4 = 209,$ which is divisible by 11 but not divisible by 2, 7 or 27.
Further, let $P(k) = 2 . 4^{2k + 1} + 3^{3k + 1}$ is divisible by 11, i.e., $2.4^{2k+1} + 3^{3k+1} = 11q$ for some integer $q$.
Now $P(k +1) = 2.4^{2k+3} +3^{3k+4}$
$= 2.4^{2k+1}.4^{2} + 3^{3k+1}.3^{3}$
$ = 16. 2.4^{2k+1} + 27.3^{3k+1}$
$=16.2.2^{2k+1}+\left(16+11\right).3^{3k+1}$
$= 16\left[2.4^{2k+1}+3^{3k+1}\right]+11.3^{3k+1}$
$=16.11q+11.3^{3k+1}$
$=11\left(16q+3^{3k+1}\right)=11m$
where $m = 16 q +3^{3k+1}$ is another integer.
$\therefore \, p(k+1) $ is divisible by 11.
$\therefore \, P(n) = 2.4^{2n+1} + 3^{3n+1}$ is divisible by 11
for all $n \in N$.