InterviewSolution
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InterviewSolution
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Let the terms a_(1),a_(2),a_(3),…a_(n) be in G.P. with common ratio r. Let S_(k) denote the sum of first k terms of this G.P.. Prove that S_(m-1)xxS_(m)=(r+1)/rSigmaSigma_(i le itj le n)a_(i)a_(j) |
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Answer» Solution :We have `a_(1)+a_(2)+…+a_(m))^(2)` `=a_(1)^(2)+a_(2)^(2)+…+a_(m)^(2)+2(a_(1)a_(2)+a_(2)a_(3)+…)` `THEREFORE[(a_(1)(1-r^(m)))/(1-r)]^(2)=(a_(1)^(2)(1-r^(2m)))/(1-r^(2))+2underset(1leiltjlen)(SigmaSigma)a_(i)a_(j)` `rArr2underset(1leiltjlen)(SigmaSigma)a_(i)a_(j)=(a_(1)^(2)(1-r^(m))^(2))/((1-r)^(2))-(a_(1)^(2)(1-r^(2m)))/(1-r^(2))` `=(a_(1)^(2)(1-r^(m)))/((1-r)^(2)(1+r))2(r-r^(m))` `=(2R)/(1+r){a_(1)cdot((1-r^(m-1)))/(1-r)}{(a_(1)(1-r^(m)))/(1-r)}` `=S_(m-1)xxS_(m)` |
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