Question

$P(n):2.7^n+3.5^n-5$ 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)

$P(n):2.7^n+3.5^n-5$ is divisible by $24.$
For $n=1,$
$P(1):2.7+3.5-5=24$, which is divisible by $24.$
Assume that $P(k)$ is true,
$i.e.2.7^k+3.5^{k+1}-5=24q$, where $q\in N...\left(i\right)$
Now, we wish to prove that $P(k+1)$ is true whenever $P(k)$ is true, i.e. $2.7^{k+1}+3.5^{k+1}-5$ is divisible by $24$. We have,
$2.7^{k+1}+3.5^{k+1}-5=2.7^k.7^1+3.5^k.5^1-5$
$=7\left[2.7^k+3.5^k-5-3.5^k+5\right]+3.5^k.5-5$
$=7\left[24q-3.5^k+5\right]+15.5^k-5<br/>=\left(7\times 24q\right)-21.5^k+35+15.5^k-5$
$=\left(7\times 24q\right)-6.5^k+30=\left(7\times 24q\right)-6\left(5^k-5\right)<br/>=\left(7\times 24q\right)-6\left(4p\right)$ [$\because (5^k-5)$ is a multiple of $4$]
$=(7\times 24q)-24p=24(7q-p)$
$=24\times r;r=7q-p$, is some natural number $...(ii)$
Thus, $P(k+1)$ is true whenever $P(k)$ is true.
Hence, by the principle of mathematical induction, $P(n)$ is true for all $n\in N.$