Permutations and Combinations

Permutations and Combinations

Question

If S1, S2 and S3 be respectively the sums of n, 2n and 3n terms of a GP., prove that S1 (S3 – S2) = (S2 – S1)2

Answer

Let a be the first term and r be the common ratio of a G.P.
                              space space space space space space space space space straight S subscript 1 space equals space Sum space of space straight n space terms space of space straight G. straight P. space equals space fraction numerator straight a left parenthesis straight r to the power of straight n minus 1 right parenthesis over denominator straight r minus 1 end fraction                            ...(i)
                                    space space straight S subscript 2 space equals space Sum space of space 2 straight n space terms space of space straight G. straight P. space equals space fraction numerator straight a left parenthesis straight r to the power of 2 straight n end exponent minus 1 right parenthesis over denominator straight r minus 1 end fraction                        ...(ii)
                                     straight S subscript 3 space equals space Sum space of space 3 straight n space terms space of space straight G. straight P. space equals space fraction numerator straight a left parenthesis straight r to the power of 3 straight n end exponent minus 1 right parenthesis over denominator straight r minus 1 end fraction                         ...(iii)
L.H.S. = straight S subscript 1 left parenthesis straight S subscript 3 minus straight S subscript 2 right parenthesis space equals space fraction numerator straight a left parenthesis straight r to the power of straight n minus 1 right parenthesis over denominator straight r minus 1 end fraction open square brackets fraction numerator straight a left parenthesis straight r to the power of 3 straight n end exponent minus 1 right parenthesis over denominator straight r minus 1 end fraction minus fraction numerator straight a left parenthesis straight r to the power of 2 straight n end exponent minus 1 right parenthesis over denominator straight r minus 1 end fraction close square brackets
         = fraction numerator straight a squared left parenthesis straight r to the power of straight n minus 1 right parenthesis over denominator left parenthesis straight r minus 1 right parenthesis squared end fraction left square bracket straight r to the power of 3 straight n end exponent minus 1 minus straight r to the power of 2 straight n end exponent plus 1 right square bracket space equals space space fraction numerator straight a squared left parenthesis straight r to the power of straight n minus 1 right parenthesis over denominator left parenthesis straight r minus 1 right parenthesis squared end fraction left square bracket straight r to the power of 3 straight n end exponent minus straight r to the power of 2 straight n end exponent right square bracket space equals space fraction numerator straight a squared straight r to the power of 2 straight n end exponent left parenthesis straight r to the power of straight n minus 1 right parenthesis squared over denominator left parenthesis straight r minus 1 right parenthesis squared end fraction
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#6 {main}</pre>
            = fraction numerator straight a squared over denominator left parenthesis straight r minus 1 right parenthesis squared end fraction left square bracket straight r to the power of 2 straight n end exponent minus straight r to the power of straight n right square bracket squared space equals space fraction numerator straight a squared straight r to the power of 2 straight n end exponent over denominator left parenthesis straight r minus 1 right parenthesis squared end fraction left parenthesis straight r to the power of straight n minus 1 right parenthesis squared

∴   L.H.S.  = R.H.S.
Hence,      straight S subscript 1 left parenthesis straight S subscript 3 minus straight S subscript 2 right parenthesis space equals space left parenthesis straight S subscript 2 minus straight S subscript 1 right parenthesis squared


          

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