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Question

a + (a + d) + (a + 2d) + ... (a + (n − 1) d) = $\frac{n}{2}\left[2a+\left(n-1\right)d\right]$

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Solution

Let P(n) be the given statement. Now, $P\left(n\right):a+\left(a+d\right)+\left(a+2d\right)+...+\left(a+\left(n-1\right)d\right)=\frac{n}{2}\left[2a+\left(n-1\right)d\right]\phantom{\rule{0ex}{0ex}}\mathrm{Step}1:\phantom{\rule{0ex}{0ex}}P\left(1\right)=a=\frac{1}{2}\left(2a+\left(1-1\right)d\right)\phantom{\rule{0ex}{0ex}}\mathrm{Hence},P\left(1\right)\mathrm{is}\mathrm{true}.\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Step}2:\phantom{\rule{0ex}{0ex}}\mathrm{Suppose}P\left(m\right)\mathrm{is}\mathrm{true}.\phantom{\rule{0ex}{0ex}}\mathrm{Then},\phantom{\rule{0ex}{0ex}}a+\left(a+d\right)+...+\left(a+\left(m-1\right)d\right)=\frac{m}{2}\left[2a+\left(m-1\right)d\right]\phantom{\rule{0ex}{0ex}}\mathrm{We}haveto\mathrm{show}\mathrm{that}P\left(m+1\right)\mathrm{is}\mathrm{true}\mathrm{whenever}P\left(m\right)\mathrm{is}\mathrm{true}.\phantom{\rule{0ex}{0ex}}Thatis,\phantom{\rule{0ex}{0ex}}a+\left(a+d\right)+...+\left(a+md\right)=\frac{\left(m+1\right)}{2}\left[2a+md\right]\phantom{\rule{0ex}{0ex}}We\mathrm{know}\mathrm{that}P\left(m\right)\mathrm{is}\mathrm{true}.\phantom{\rule{0ex}{0ex}}Thus,wehave:\phantom{\rule{0ex}{0ex}}a+\left(a+d\right)+...+\left(a+\left(m-1\right)d\right)=\frac{m}{2}\left[2a+\left(m-1\right)d\right]\phantom{\rule{0ex}{0ex}}⇒a+\left(a+d\right)+...+\left(a+\left(m-1\right)d\right)+\left(a+md\right)=\frac{m}{2}\left[2a+\left(m-1\right)d\right]+\left(a+md\right)\left[\mathrm{Adding}\left(a+md\right)\mathrm{to}\mathrm{both}\mathrm{sides}\right]\phantom{\rule{0ex}{0ex}}⇒P\left(m+1\right)=\frac{1}{2}\left[2am+{m}^{2}d-md+2a+2md\right]\phantom{\rule{0ex}{0ex}}⇒P\left(m+1\right)=\frac{1}{2}\left[2a\left(m+1\right)+md\left(m+1\right)\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left(m+1\right)\left(2a+md\right)\phantom{\rule{0ex}{0ex}}Thus,P\left(m+1\right)\mathrm{is}\mathrm{true}.\phantom{\rule{0ex}{0ex}}Bythep\mathrm{rinciple}\mathrm{of}m\mathrm{athematical}i\mathrm{nduction},P\left(n\right)\mathrm{is}\mathrm{true}\mathrm{for}\mathrm{all}n\in N.$

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