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Use binomial theorem to show that $7^n + 2$ is divisible by $3$.

What I've done: $$\begin{aligned} (a+b)^n &= \sum_{r=0}^{n}\dbinom{n}{r}a^{n-r}b^r \\\\ 7^n+2 &=( 1+6)^n+2\\ &= \sum_{r=0}^{n}\dbinom{n}{r}1^{n-r}\times6^r+2 \\ &= \sum_{r=0}^{n}\dbinom{n}{r}6^r+2 \end{aligned}$$

But I'm not sure what to do from there. Would appreciate any help.

Soumik Mukherjee
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zava
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    Your binormial formula is incorrect – Arctic Char Apr 01 '23 at 09:03
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    Use the fact that $\binom n0r^0+2=3$. – José Carlos Santos Apr 01 '23 at 09:10
  • Either use $3\mid 7\!-\!1\mid 7^n\!-1$ by the [Factor Theorem](https://math.stackexchange.com/a/1844838/242) in the first dupe (where [this answer](https://math.stackexchange.com/a/17809/242) proves it via BT=Binomial Theorem), or directly invoke BT as in the 2nd dupe. Much easier, use the Congruence Power Rule [as here](https://math.stackexchange.com/a/17800/242) in the 1st dupe. – Bill Dubuque Apr 01 '23 at 20:04

1 Answers1

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$$7^n + 2 = (6+1)^n + 2$$
Apply the binomial theorem to evaluate the first term on the RHS

$$\left(\sum_\limits{k=0}^n {n\choose k} 6^{k}1^{n-k}\right) + 2$$

In that summation every term where $k > 0$ we see that $3|6^{k}$ and hence $3\Bigg|\displaystyle{n\choose k} 6^{k}1^{n-k}$

When $k = 0$ we have $\displaystyle{n\choose 0} 6^{0}1^{n} = 1$

And $3|1+2$

Deeponjit Bose
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user317176
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  • I proposed a duplicate but there are plenty others: may be you could propose a better one (instead of this answer). – Anne Bauval Apr 01 '23 at 09:32
  • Please strive not to post more (dupe) answers to dupes of FAQs, cf. recent site policy announcement [here](https://math.meta.stackexchange.com/q/33508/242). – Bill Dubuque Apr 01 '23 at 20:00
  • @BillDubuque this question was not marked as dupe when I answered it. However, I did not search the database for similar questions before I posted my answer. – user317176 Apr 01 '23 at 20:30
  • Surely you have enough experience by now to know questions like this have been answered *many* times already. If you are unable help organize the site then please refrain from answering such dupes (cf. linked site policy). – Bill Dubuque Apr 01 '23 at 20:34