Verified Answer :

12+23+3*4+...+n(n+1)=
(1²+1)+(2²+2)+(3²+3)+...+(n²+n)=
(1² + 2² + 3² + ... + n²) + (1+2+3+...+n)=
\(\frac{n(n+1)(2n+1)}{6} * \frac{n(n+1)}{2}\) =
\(\frac{n(n+1)(n+2)}{3}\)

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