Good use of induction!!!

This is related to the Previous note about Induction

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

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

Problem is to prove that 24(27n+35n5)24 \mid (2\cdot 7^n+3\cdot 5^n -5) ...... nN\forall n\in \mathbb{N}

Observe that for n=1n=1, we have 27n+35n5=14+155=242\cdot 7^n+3\cdot 5^n -5 = 14+15-5=24. Hence the result is true for 11.

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

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

Consider the term 27k+1+35k+152\cdot 7^{k+1} +3\cdot 5^{k+1} -5.

This term is equivalent to 7(27k+35k5)65k+307(2\cdot7^k +3\cdot 5^k -5) - 6\cdot 5^k +30 ...... this is because 7×35k65k=(216)5k=35k+17\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=kn=k, hence in the term, i.e.

27k+1+35k+15=7(27k+35k5)divisible by 24 by assumption65k+30\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 65k+30-6\cdot 5^k +30 is divisible by 24, see that

65k+30=6(5k5)-6\cdot 5^k+30= -6(5^k-5). So the problem has reduced to proving 246(5k5)24\mid -6(5^k -5) or even simpler, 45k54 \mid 5^k-5

We know that 51(mod4)    5k1(mod4)5 \equiv 1\pmod{4}\implies 5^k \equiv 1 \pmod{4}

And 5k51(mod4)    5k5110(mod4)5^k\equiv 5\equiv 1 \pmod{4} \implies 5^k-5\equiv 1-1\equiv 0 \pmod{4}

Hence 45k54 | 5^k-5 . Hence the other term in the expression is also divisible by 24, thus if the result is assumedassumed to be true for n=kn=k then we are obtainingobtaining it to be true for n=k+1n=k+1 and for n=1n=1 we already proved, thus it's true for ALLALL natural numbers.

Note by Aditya Raut
6 years, 10 months ago

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Aamir Faisal Ansari - 6 years, 10 months ago

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Good one

Usama Khidir - 6 years, 9 months ago

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