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  • Find the sum to n terms of each of the series 3 × 1^2 + 5 × 2^2 + 7 × 3^2 + ...

3.   Find the sum to n terms of each of the series  3 \times 1 ^ 2 + 5 \times 2 ^ 2 + 7 \times +....+ 20 ^ 2

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best_answer

the series  3 \times 1 ^ 2 + 5 \times 2 ^ 2 + 7 \times +....+ 20 ^ 2

nth term  = (2n+1)(n^2)=2n^3+n^2=a_n

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

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

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

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

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

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

                   =\left [ \frac{n(n+1)}{2} \right ]\frac{(3n^2+5n+1)}{3}

                  = \frac{n(n+1)(3n^2+5n+1)}{6}   

Thus, the sum is 

                       = \frac{n(n+1)(3n^2+5n+1)}{6}

Posted by

seema garhwal

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