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Q.
$P\left(n\right): 2\cdot7^{n} + 3\cdot5^{n} - 5 $ is divisible by
Principle of Mathematical Induction
Solution:
$P\left(n\right) : 2\cdot7^{n} +3\cdot5^{n}-5$ is divisible by $24$.
For $n =1, P\left(1\right) : 2\cdot7 +3\cdot5 -5 =24$, is divisible by $24$.
Assume that $P\left(k\right)$ is true
i.e. $2\cdot7^{k} +3\cdot5^{k} -5 = 24q$, where $q\in N \quad...\left(i\right)$
Now, we wish to prove that $P\left(k + 1\right)$ is true whenever $P\left(k\right)$
is true, i.e $2\cdot7^{k+1} +3\cdot5^{k+1} -5$ is divisible by $24$.
We have
$ 2\cdot7^{k+1} +3\cdot5^{k+1}-5 =2\cdot7^{k} \cdot7^{1} +3\cdot5^{k} \cdot5^{1} -5 $
$=7\left[2\cdot7^{k} +3\cdot5^{k}-5-3\cdot5^{k}+5\right] + 3\cdot5^{k} \cdot5 -5$
$= 7\left[24q -3\cdot5^{k}+5\right]+15\cdot5^{k}-5$
$=\left(7\times24q\right)-6\cdot5^{k}+30=\left(7\times24q\right)-6\left(5^{k}-5\right)$
$= \left(7\times24q\right) - 6 \left(4p\right) \quad\left[\because\left(5^{k} -5\right) {\text{ is multiple of }} 4\right]$
$ =\left(7\times24q\right)-24p=24\left(7q-p\right)$
$ = 24\times r : r =7q-p$, is some natural number. $\quad ...\left(ii\right)$
Thus $P\left(k + 1\right)$ is true whenever $P\left(k\right)$ is true.
Hence, by the principle of mathematical induction, $P\left(n\right)$ is true for all $n\in N$.