Browse Questions

# Prove that  $2.7^n+3.5^n-5$ is divisible by 24, for all $n \in N$.

Let the statement $P(n)$ be defined as
$P(n) : 2.7^n+3.5^n-5$ is divisible by 24
We note that $P(n)$ is true when $n=1$, since $2.7+3.5-5=24$, which is divisible by 24.
Assume that $P(k)$ is true.
i.e., $2.7^k+3.5^k-5=24q$ when $q \in N$------------(1)
Now, we wish to prove that $P(k+1)$ is true whenever $P(k)$ is true.
We have
$2.7^{k+1}+3.5^{k+1}-5=2.7^k.7^1+3.5^k.5^1-5$
$=7[2.7^k+3.5^k-5-3.5^k+5]+3.5^k.5-5$
$=7[24q-3.5^k+5]+15.5^k-5$
$= 7 \times 24q-21.5^k+35+15.5^k-5$
$= 7 \times 24q - 6.5^k+30$
$= 7 \times 24q - 6(5^k-5)$
$= 7 \times 24q-6(4p) [ (5^k-5)$ is a multiple of 4 (why ?) ]
$= 7 \times 24q-24p$
$= 24(7q-p)$
$= 24 \times r; r =7q-p$, is some natural number-----------(2)
The expression on the R.H.S of (1) is divisible by 24. Thus $P(k+1)$ is true whenever $P(k)$ is true.
Hence, by principle of mathematical induction , $P(n)$ is true for all $n \in N$.