Prove by Induction!

Show that $2\cdot7^{n} + 3\cdot5^{n} -5$ is divisible by 24. Note by Swapnil Das
4 years ago

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Step $1$: when $n=1$, we have $2 \cdot 7^{n}+3 \cdot 5^{n}-5=2 \cdot 7+3 \cdot 5 -5=24$ which is divisible by $24$.

Step $2$: Assume this to be true for $n=m$. Then $2 \cdot 7^{m}+3 \cdot 5^{m}-5$ is divisible by $24$.

Step $3$: Now, $2 \cdot 7^{m+1}+3 \cdot 5^{m+1}-5$

$=2 \cdot 7^{m} \times 7+ 3 \cdot 5^{m} \times 5-5$

$=14 \cdot 7^{m}+15 \cdot 5^{m}-5$

$=2 \cdot 7^{m}+3 \cdot 5^{m}-5+12 \cdot 7^{m}+12 \cdot 5^{m}$

Since $2 \cdot 7^{m}+3 \cdot 5^{m}-5$ is divisible by $24$, we have only to prove that $12 \cdot 7^{m}+12 \cdot 5^{m}$ is divisible by $24$.

$12 \cdot 7^{m}+12 \cdot 5^{m}=12(7^{m}+5^{m})$ which is divisible by $24$ since $7^{m}$ and $5^{m}$ are both odd numbers and their sum is an even number.

Hence $2 \cdot 7^{n}+3 \cdot 5^{n}-5$ is divisible by $24$ for all integers $n>0$.

Excellent, thank you genius!

- 4 years ago

Hint: Show that it is true for n=1. Then show that if the statement is true for any arbitrary positive integer m, then it is true for m+1.

That's an obvious hint lol xD

- 4 years ago

Can you show the m+1 step? This is where I'm getting stuck at.

- 4 years ago

Only tool (with me ):MATHEMATICAL INDUCTION

- 4 years ago

I'm surprised that you don't know Induction. You should read the wiki and be familiar with it. This is a standard problem in divisibility by induction.

Staff - 4 years ago

Lol, I assumed powers of $5$ to be even by mistake, and thus couldn't get the result!

- 4 years ago

This is true only for n equals 1. It is not true for numbers bigger than 1.

- 3 years, 11 months ago

@ Mohamed Sultan - You are wrong. Go back and try it for n = 2, 3, and 4, for instance.

- 1 year, 2 months ago