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  • Find the sum to n terms of each of the series in 1^2 + (1^2 + 2^2) + (1^2 + 2^2 + 3^2) +...

7. Find the sum to n terms of each of the series in 1 ^ 2 + ( 1 ^2 +2 ^ 2 ) + ( 1 ^ 2 +2 ^ 2 + 3 ^ 2 ) ...

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best_answer

series =    1 ^ 2 + ( 1 ^2 +2 ^ 2 ) + ( 1 ^ 2 +2 ^ 2 + 3 ^ 2 ) ...

n th term  = a_n=1^2+2^2+3^2+...................n^2=\frac{n(n+1)(2n+1)}{6}

                                                                                  =\frac{n(2n^2+3n+1)}{6}=\frac{2n^3+3n^2+n}{6}

S_n=\sum _{k=1}^{n} a_k=\sum _{k=1}^{n} \frac{2k^3+3k^2+k}{6}

                        =\frac{1}{3}\sum _{k=1}^{n} k^3+\frac{1}{2}\sum _{k=1}^{n} k^2+\frac{1}{6}\sum _{k=1}^{n} k

                     =\frac{1}{3}\left [ \frac{n(n+1)}{2} \right ]^2+\frac{1}{2}.\frac{n(n+1)(2n+1)}{6}+\frac{1}{6}\frac{n(n+1)}{2}

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

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

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

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

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

           =\left [ \frac{n(n+1)^2(n+2)}{12} \right ] 

Posted by

seema garhwal

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