Prove the following by the principle of mathematical induction:`\ 2. 7^n+3. 5^n-5` is divisible 25 for all `n in Ndot`

Explanation

Let ` P(n) : (2*7^(n)+3*5^(n)-5)` is divisible by 24.
For n = 1 , the given expression becomes `(2*7^(1)+3*5^(1)-5) = 24`, which is clearly divisible by 24.
So, the given statement is true for n =1, i.e., P(1) is true.
Let P(k) be true. Then,
`P(k): (2*7^(k)+3*5^(k)-5)` is divisible by 24
` rArr" " (2*7^(k)+3*5^(k) -5) = 24m`, for some ` m in N`. ...(i)
Now, `(2*7^(k+1)+3*5^(k+1)-5)`
` = (2*7^(k) *7+3*5^(k)*5-5)`
` = 7(2*7^(k)+3*5^(k)-5) - 6*5^(k) +30`
` =(7xx 24m)-6(5^(k)-5)" "` [using (i)]
` = (24xx 7m)-6xx5xx(5^(k-1)-1)`
` =(24xx7m)-5xx24p[:' (5^(k-1)-1)" is divisible by " (5-1), i.e., 4 rArr (5^(k-1)-1) = 4p]`
` = 24 xx(7m-5p)`, which is clearly divisible by 24.
` :. P(k+1): (2*7^(k+1)+3*5^(k+1)-5)` is divisible by 24.
Thus , P(k+1) is true, whenever P(k) is true.
` :. ` P(1) is true and P(k+1) is true, whenever P(k) is true.
Hence, by the principle of mathematical induction, it follows that ` (2*7^(n)+3*5^(n)-5)` is divisible by 24 for all ` n in N`.