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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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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. – 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