Question

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

Answer 1

$2(7{)}^m+3(5{)}^m-5=P\left(n\right)$

divisible by $24=4\times$natural no

for $n=1$ $P\left(n\right)=$ true for $n=1$

$P(k+1)={27}^{k+1}+{35}^{k+1}-5$

$\Rightarrow 2.2\times 7^k+{\mathrm{5.3.5}}^k-5=24m$

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

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

$(5^k-5)$ is a multiple of $4$

$=7\times 24m-6\left(4p\right)$

$P$ is a natural No

$=7\times 24m-24P$

$=24(7m-P)=24\times r$

$r=7m=P$ is some natural No $P(k+1)$ is true whenever $P\left(k\right)$ is true.