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 \(\color{Blue}{TRUE}\) for \(n=k\), and we'll prove for \(n=k+1\)

For this, we use a \(\color{Green}{\text{little manipulation}}\).

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
4 years, 10 months ago

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Aamir Faisal Ansari - 4 years, 10 months ago

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Good one

Usama Khidir - 4 years, 8 months ago

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