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24.  If S_1 , S_2 , S_3 are the sum of first n natural numbers, their squares and their cubes, respectively, show that 9 S ^2 _2 = S_3 ( 1+ 8 S_1)

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To prove : 9 S ^2 _2 = S_3 ( 1+ 8 S_1)

From given information,

             S_1=\frac{n(n+1)}{2}

          S_3=\frac{n^2(n+1)^2}{4}

Here ,RHS= S_3 ( 1+ 8 S_1)

\Rightarrow S_3 ( 1+ 8 S_1)=\frac{n^2(n+1)^2}{4}\left ( 1+8\frac{n(n+1)}{2} \right )

                              =\frac{n^2(n+1)^2}{4}\left ( 1+4.n(n+1) \right )

                              =\frac{n^2(n+1)^2}{4}\left ( 1+4n^2+4n \right )

                          =\frac{n^2(n+1)^2}{4}\left ( 2n+1 \right )^2

                           =\frac{n^2(n+1)^2(2n+1)^2}{4}.........................................................1

Also, RHS=9S_2^2

\Rightarrow 9S_2^2=\frac{9\left [ n(n+2)(2n+1) \right ]^2}{6^2}

               =\frac{9\left [ n(n+2)(2n+1) \right ]^2}{36}

               =\frac{\left [ n(n+2)(2n+1) \right ]^2}{4}..........................................2

From equation 1 and 2 , we have 

        9S_2^2=S_3(1+8S_1)=\frac{\left [ n(n+2)(2n+1) \right ]^2}{4}

Hence proved .

Posted by

seema garhwal

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