# Good use of induction!!!

This is related to the Previous note about Induction

Out of the problems stated in the note, $(a)$ to $(d)$ and $(f)$ are standard results, you can use them like identities.

Here's the proof of $(e)$ which is slightly different than other ones.

Problem is to prove that $24 \mid (2\cdot 7^n+3\cdot 5^n -5)$ ...... $\forall n\in \mathbb{N}$

Observe that for $n=1$, we have $2\cdot 7^n+3\cdot 5^n -5 = 14+15-5=24$. Hence the result is true for $1$.

Assume that the result is $\color{#3D99F6}{TRUE}$ for $n=k$, and we'll prove for $n=k+1$

For this, we use a $\color{#20A900}{\text{little manipulation}}$.

Consider the term $2\cdot 7^{k+1} +3\cdot 5^{k+1} -5$.

This term is equivalent to $7(2\cdot7^k +3\cdot 5^k -5) - 6\cdot 5^k +30$ ...... this is because $7\times 3 \cdot 5^k - 6\cdot 5^k = (21-6)5^k = 3\cdot 5^{k+1}$ and this term we want in our proof!!!

Returning to the proof, we have assumed that the result is true for $n=k$, hence in the term, i.e.

$\displaystyle 2\cdot 7^{k+1} +3\cdot 5^{k+1} -5 =7 \underbrace{(2\cdot 7^k +3\cdot 5^k -5)}_{\text{divisible by 24 by assumption}} - 6\cdot 5^k +30$

Thus what only remains to prove is $-6\cdot 5^k +30$ is divisible by 24, see that

$-6\cdot 5^k+30= -6(5^k-5)$. So the problem has reduced to proving $24\mid -6(5^k -5)$ or even simpler, $4 \mid 5^k-5$

We know that $5 \equiv 1\pmod{4}\implies 5^k \equiv 1 \pmod{4}$

And $5^k\equiv 5\equiv 1 \pmod{4} \implies 5^k-5\equiv 1-1\equiv 0 \pmod{4}$

Hence $4 | 5^k-5$ . Hence the other term in the expression is also divisible by 24, thus if the result is $assumed$ to be true for $n=k$ then we are $obtaining$ it to be true for $n=k+1$ and for $n=1$ we already proved, thus it's true for $ALL$ natural numbers. Note by Aditya Raut
6 years, 10 months ago

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## Comments

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Nice.

- 6 years, 10 months ago

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Good one

- 6 years, 9 months ago

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