Question

${2.7}^n+{3.5}^n-5$ is divisible by 24 for all $nϵN$.

Answer 1

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

$Forn=1$

$2.7+3.5-5=24$

It is divisible of 24

$\Rightarrow$ P(n) is true for n = 1

Let P(n) is true for n = k, so

${2.7}^k+{3.5}^k-5$ is divisible by 24

${2.7}^k+{3.5}^k-5=24\lambda$ .......(1)

We have to show that,

$={2.7}^{(k+1)}+{3.5}^{(k+1)}-5$

$={2.7}^k.7+{3.5}^k.5-5$

$=(24\lambda -{3.5}^k+5)7+{15.5}^k-5$

$=24.7\lambda -{21.5}^k+35+{15.5}^k-5$

$=24.7\lambda -{6.5}^k+30$

$=24.7\lambda -6(5^k-5)$

$=24.7\lambda -6\left(20v\right)$

[Since $5^k-5$ is multiple of 20]

$=24(7\lambda -5v)$

$=24\mu$

$\Rightarrow$ P(n) is true for n = k + 1

$\Rightarrow$ P(n) is true for all $nϵN$ by PMI