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| 1. |
a+ar+ar*2+....+ar*n-1=a(r*n-1)/r-1 |
| Answer» Let P(n) = a + ar + ar2 + ..... + arn-1{tex} = \\frac{{a({r^n} - 1)}}{{r - 1}}{/tex}.For n = 1{tex}P(1) = a{r^{1 - 1}} = \\frac{{a({r^1} - 1)}}{{r - 1}} \\Rightarrow a = a{/tex}{tex}\\therefore {/tex}\xa0P(1) is trueLet P(n) be true for n = k{tex}\\therefore P(k) = a + ar + a{r^2} + .... + a{r^{k - 1}}{/tex}{tex} = \\frac{{a({r^k} - 1)}}{{r - 1}}{/tex}\xa0.... (i)For n = k + 1R.H.S. {tex} = \\frac{{a({r^{k + 1}} - 1)}}{{r - 1}}{/tex}L.H.S. {tex} = \\frac{{a({r^k} - 1)}}{{r - 1}} + a{r^k}{/tex}\xa0[ Using (i)]{tex} = \\frac{{a{r^k}}}{{r - 1}} - \\frac{a}{{r - 1}} + a{r^k}{/tex}{tex} = a{r^k} \\cdot \\left( {\\frac{1}{{r - 1}}+1} \\right) - \\frac{a}{{r - 1}} = \\frac{{a{r^{k + 1}}}}{{r - 1}} - \\frac{a}{{r - 1}}{/tex}{tex} = \\frac{{a{r^{k + 1}} - a}}{{r - 1}}{/tex}{tex} = \\frac{{a({r^{k + 1}} - 1)}}{{r - 1}}{/tex}{tex}\\therefore {/tex}\xa0P(k + 1) is trueThus P(k) is true {tex} \\Rightarrow {/tex}\xa0P(k + 1) is trueHence by principle of mathematical induction, P(n) is true for all {tex}n \\in N{/tex}. | |