Question

$P\left(n\right):2\cdot 7^n+3\cdot 5^n-5$ is divisible by

• A. $24,\forall n\in N$
• B. $21,\forall n\in N$
• C. $35,\forall n\in N$
• D. $50,\forall n\in N$

Answer 1

Answer: (A)

$P\left(n\right):2\cdot 7^n+3\cdot 5^n-5$ is divisible by $24$.
For $n=1,P\left(1\right):2\cdot 7+3\cdot 5-5=24$, is divisible by $24$.
Assume that $P\left(k\right)$ is true
i.e. $2\cdot 7^k+3\cdot 5^k-5=24q$, where $q\in N...\left(i\right)$
Now, we wish to prove that $P\left(k+1\right)$ is true whenever $P\left(k\right)$
is true, i.e $2\cdot 7^{k+1}+3\cdot 5^{k+1}-5$ is divisible by $24$.
We have
$2\cdot 7^{k+1}+3\cdot 5^{k+1}-5=2\cdot 7^k\cdot 7^1+3\cdot 5^k\cdot 5^1-5$
$=7\left[2\cdot 7^k+3\cdot 5^k-5-3\cdot 5^k+5\right]+3\cdot 5^k\cdot 5-5$
$=7\left[24q-3\cdot 5^k+5\right]+15\cdot 5^k-5$
$=\left(7\times 24q\right)-6\cdot 5^k+30=\left(7\times 24q\right)-6\left(5^k-5\right)$
$=\left(7\times 24q\right)-6\left(4p\right)\left[\because \left(5^k-5\right)\text{ is multiple of }4\right]$
$=\left(7\times 24q\right)-24p=24\left(7q-p\right)$
$=24\times r:r=7q-p$, is some natural number. $...\left(ii\right)$
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 $n\in N$.