Question

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

Answer 1

Let $P\left(n\right):{2.7}^n+{3.5}^n-5$ is divisible by $24$

We note that $P\left(n\right)$ is true when $n=1$, since $2.7+3.5-5=24$. which is divisible by $24$.

Assume that $P\left(k\right)$ is true.

i.e. ${2.7}^k+{3.5}^k-5=24q$ when $q\in N$ -------------- ( 1 )

Now, we have to prove that $P(k+1)$ is true whenever $P\left(k\right)$ is true.

We have

${2.7}^{k+1}+{3.5}^{k+1}-5$

$\Rightarrow$ ${2.7}^k{.7}^1+{3.5}^k{.5}^1-5$

$\Rightarrow$ $7[{2.7}^k+{3.5}^k-5-{3.5}^k+5]+{3.5}^k.5-5$

$\Rightarrow$ $7[24q-{3.5}^k+5]+{15.5}^k-5$

$\Rightarrow$ $2\times 24q-{21.5}^k+35+{15.5}^k-5$

$\Rightarrow$ $7\times 24q-{6.5}^k+30$

$\Rightarrow$ $7\times 24q-6(5^k-5)$

$\Rightarrow$ $7\times 24q-6\left(4p\right)$ [ $(5^k-5)$ is multiple of $4$ ]

$\Rightarrow$ $7\times 24q-24p$

$\Rightarrow$ $24(7p-q)$

$\Rightarrow$ $24\times r;r=7p-q.$ is some natural number ---------- ( 2 )

The expression on the $R.H.S$ oof ( 1 ) is divisible by $24$. Thus $P(k+1)$ is true whenever $P\left(k\right)$ is true.

Hence, by principle of mathematical induction , $P\left(n\right)$ is true for all $n\in N.$