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To n th term is n.(n+1).(n+2)

t_n = n³ + 3n² + 2n

so S_n for n =  \frac{n(n+1)}{2}

S_n for n² =   \frac{n(n+1)(2n+1)}{6}

S_n for n³ =  {\frac{n(n+1)}{2}}^2

so  t_n = n³ + 3n² + 2n

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

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

 n(n+1)[ \frac{n(n+1)}{4}+\frac{2n+1}{2}+1]


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