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Question

# Using binomial theorem, prove that ${2}^{3n}-7n-1$ is divisible by 49, where $n\in N$.

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Solution

## ${2}^{3n}-7n-1={8}^{n}-7n-1$ ...(1) $\mathrm{Now},\phantom{\rule{0ex}{0ex}}{8}^{n}=\left(1+7{\right)}^{n}\phantom{\rule{0ex}{0ex}}={}^{n}C_{0}+{}^{n}C_{1}×{7}^{1}+{}^{n}C_{2}×{7}^{2}+{}^{n}C_{3}×{7}^{3}+{}^{n}C_{4}×{7}^{4}+...+{}^{n}C_{n}×{7}^{n}\phantom{\rule{0ex}{0ex}}⇒{8}^{n}=1+7n+49\left[{}^{n}C_{2}+{}^{n}C_{3}×{7}^{1}+{}^{n}C_{4}×{7}^{2}+...+{}^{n}C_{n}×{7}^{n-2}\right]\phantom{\rule{0ex}{0ex}}⇒{8}^{n}-1-7n=49×\left(\mathrm{An}\mathrm{integer}\right)\phantom{\rule{0ex}{0ex}}\mathrm{Now},\phantom{\rule{0ex}{0ex}}{8}^{n}-1-7n\mathrm{is}\mathrm{divisible}\mathrm{by}49\phantom{\rule{0ex}{0ex}}\mathrm{Or},\phantom{\rule{0ex}{0ex}}{2}^{3n}-1-7n\mathrm{is}\mathrm{divisible}\mathrm{by}49\left[\mathrm{From}\left(1\right)\right]$

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