Answer :
1*2+2*3+3*4+...+n(n+1)=
(1²+1)+(2²+2)+(3²+3)+...+(n²+n)=
(1² + 2² + 3² + ... + n²) + (1+2+3+...+n)=
[tex] \frac{n(n+1)(2n+1)}{6} * \frac{n(n+1)}{2} [/tex] =
[tex] \frac{n(n+1)(n+2)}{3} [/tex]
(1²+1)+(2²+2)+(3²+3)+...+(n²+n)=
(1² + 2² + 3² + ... + n²) + (1+2+3+...+n)=
[tex] \frac{n(n+1)(2n+1)}{6} * \frac{n(n+1)}{2} [/tex] =
[tex] \frac{n(n+1)(n+2)}{3} [/tex]
