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Question

# 72n + 23n−3. 3n−1 is divisible by 25 for all n ∈ N.

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Solution

## Let P(n) be the given statement. Now, $P\left(n\right):{7}^{2n}+{2}^{3n-3}.{3}^{n-1}\mathrm{is}\mathrm{divisible}\mathrm{by}25.\phantom{\rule{0ex}{0ex}}\mathrm{Step}1:\phantom{\rule{0ex}{0ex}}P\left(1\right):{7}^{2}+{2}^{3-3}.{3}^{1-1}=49+1=50\phantom{\rule{0ex}{0ex}}It\mathrm{is}\mathrm{divisible}\mathrm{by}25.\phantom{\rule{0ex}{0ex}}\mathrm{Thus},P\left(1\right)\mathrm{is}\mathrm{true}\phantom{\rule{0ex}{0ex}}\mathrm{Step}2:\mathrm{Let}P\left(m\right)\mathrm{be}\mathrm{true}.\phantom{\rule{0ex}{0ex}}Now,\phantom{\rule{0ex}{0ex}}{7}^{2m}+{2}^{3m-3}.{3}^{m-1}\mathrm{is}\mathrm{divisible}\mathrm{by}25.\phantom{\rule{0ex}{0ex}}\mathrm{Suppose}:\phantom{\rule{0ex}{0ex}}{7}^{2m}+{2}^{3m-3}.{3}^{m-1}=25\lambda ...\left(1\right)\phantom{\rule{0ex}{0ex}}\mathrm{We}\mathrm{have}\mathrm{to}\mathrm{show}\mathrm{that}P\left(m+1\right)\mathrm{is}\mathrm{true}\mathrm{whenever}P\left(m\right)\mathrm{is}\mathrm{true}.\phantom{\rule{0ex}{0ex}}\mathrm{Now},\phantom{\rule{0ex}{0ex}}P\left(m+1\right)={7}^{2m+2}+{2}^{3m}.{3}^{m}\phantom{\rule{0ex}{0ex}}={7}^{2m+2}+{7}^{2}.{2}^{3m-3}.{3}^{m-1}-{7}^{2}.{2}^{3m-3}.{3}^{m-1}+{2}^{3m}.{3}^{m}\phantom{\rule{0ex}{0ex}}={7}^{2}\left({7}^{2m}+{2}^{3m-3}.{3}^{m-1}\right)+{2}^{3m}.{3}^{m}\left(1-\frac{49}{24}\right)\phantom{\rule{0ex}{0ex}}={7}^{2}×25\lambda -{2}^{3m}.{3}^{m}×\frac{25}{{2}^{3}.{3}^{1}}\left[\mathrm{Using}\left(1\right)\right]\phantom{\rule{0ex}{0ex}}=25\left(49\lambda -{2}^{3m-3}.{3}^{m-1}\right)\phantom{\rule{0ex}{0ex}}It\mathrm{is}\mathrm{divisible}\mathrm{by}25.\phantom{\rule{0ex}{0ex}}\mathrm{Thus},P\left(m+1\right)\mathrm{is}\mathrm{true}.\phantom{\rule{0ex}{0ex}}Bythep\mathrm{rinciple}\mathrm{of}m\mathrm{athematical}i\mathrm{nduction},P\left(n\right)\mathrm{is}\mathrm{true}\mathrm{for}\mathrm{all}n\in N.$

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