Given that $f(n)= 7^{2n} + (2^{3n-3})(3^{n-1})$, show that $f(n+1)-24f(n)=25(7^{2n})$

Question

Given that $f(n)= 7^{2n} + (2^{3n-3})(3^{n-1})$, show that $$f(n+1)-24f(n)=25(7^{2n}).$$

Hence, use the method of induction of prove that $f(n)$ is divisible by $25$, where $n$ is any positive integer.

I don't quite understand how one would start, any advice would be greatly appreciated...

Answer 1

First part:

Given: $f(n)= 7^{2n} + (2^{3n-3})(3^{n-1})$

To prove: $f(n+1)-24f(n)=25(7^{2n})$

Proof:

$$f(n+1)-24f(n)= 7^{2n+2} + (2^{3n})(3^{n})-24(7^{2n}) - 24 (2^{3n-3})(3^{n-1})$$ $$\qquad\qquad\qquad\quad= 49(7^{2n}) + (2^{3n})(3^{n})- 24(7^{2n})- 24 (2^{3n})(3^{n})$$ $$= \underline{25(7^{2n})}$$

Second part:

To prove: $f(n)= 7^{2n} + (2^{3n-3})(3^{n-1})$ is divisible by $25$ for $n\in\Bbb Z$

Proof:

① For $n = 1$:

$$f(1) = 7^2 + (2^0)(3^0)$$ $$= 49+1$$ $$= 50$$

$$\therefore \text{Statement is true for n = 1. As the above is in fact divisible by 25}$$

② For $n = k$:

$$ \text{Assume statement is true for } n = k$$ $$ \text{That is, } f(k) = 25p \text{ where } p\in\Bbb Z$$

③ For $n = k+1$:

$$ \text{We now assume statement is true for } n = k+1$$ $$ \text{That is, } f(k+1) = 25q \text{ where } q\in\Bbb Z$$

Now, using the result proved above:

$$f(k+1) - 24f(k) = 25(7^{2k})$$ $$= 24(25p)+25(7^{2k})$$ $$= 25(24p+7^{2k})$$ $$= 25q$$

So, if the statement is true for $n = k$, it must also be true for $n = k+1$

But statement is true for $n = 1$.

$$ \text{Hence, by induction, statement is true for all $n\in\Bbb Z^+$}$$

Answer 2

hint

Check that $$f(n)=49^n+24^{n-1}$$

then

$$f(n+1)=49^{n+1}+24^n$$ and

$$f(n+1)-24f(n)=$$ $$49.49^n+24^n-24.49^n-24^n=$$ $$25.49^n=...$$