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Question

If 0<r<1 and n is a +ive integer then prove that (2n+1)rn(1r)<1r2n+1.

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Solution

We have to prove that 1r2n+11r>(2n+1)r2
L.H.S. = Sum of a G.P. of 2n+1 terms whose first terms is 1, and common ratio r.
L.H.S. =1+r+r2+...+r2n
=(1+r2n)+(r+r2n1)+(r2+r2n2)+...n pairs +rn
Apply A.M.>G.M. on each pair
L.H.S.>(2rn+2rn+...+2rnn terms)+rn
=2n.rn+rn=(2n+1)rn

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