Question

Prove that $2\cdot 7^n+3\cdot 5^n-5$ is divisible by $24$, for all $n\in N$.

Answer 1

Step $\left(1\right):$ Assume given statement
Let the statement $P\left(n\right)$ be defined as
$P\left(n\right):2\cdot 7^n+3\cdot 5^n-5$ is divisible by $24$.

Step $\left(2\right):$ Checking statement $P\left(n\right)$ for $n=1$
Put $n=1$ in $P\left(n\right)$, we get
$2\cdot 7^1+3\cdot 5^1-5$
$=14+15-5$
$=24$ which is divisible by $24$
Thus, $P\left(n\right)$ is true for $n=1$.

Step $\left(3\right):$ $P\left(n\right)$ for $n=K$
Put $n=K$ in $P\left(n\right)$, and assume that $P\left(K\right)$ is true for some natural number $K$ i.e.
$P\left(K\right):2\cdot 7^K+3\cdot 5^K-5=24q$, where $q\in N\cdots \left(1\right)$

Step $\left(4\right):$ Checking statement $P\left(n\right)$ for $n=K+1$
Now, we wish to prove that $P(K+1)$ is true whenever $P\left(K\right)$ is true, we have
$2\cdot 7^{K+1}+3\cdot 5^{K+1}-5$
$=2\cdot 7\cdot 7^K+3\cdot 5\cdot 5^K-5$
$=7[2\cdot 7^K+3\cdot 5^K-5-3\cdot 5^K+5]+15\cdot 5^K-5$
$=7[24q-{3.5}^K+5]+{15.5}^K-5$ (Using $\left(1\right)$)
$=7\times 24q-21\cdot 5^K+35+15\cdot 5^K-5$
$=7\times 24q+5^K(15-21)+35-5$
$=7\times 24q-{6.5}^K+30$
$=7\times 24q-6(5^K-5)$
$=7\times 24q-6\left(4p\right)$
[$\because (5^K-5)$ is a multiple of $4$]
$=7\times 24q-24p$
$=24(7q-p)$
$=24\times r$ where $r=7q-p$, is some natural number.
The expression is divisible by $24$.
Thus $P(K+1)$ is true whenever $P\left(K\right)$ is true.
Final Answer:
Hence by principle of mathematical induction, $P\left(n\right)$ is true for all $n\in N$.