Question

1•2•3+2•3•4+4•5•6+….n(n+1)(n+2)

Answer 1

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Answer 2

$\color{red}S\color{w}= 1 \cdot 2 \cdot 3 +2 \cdot ... + n(n + 1)(n + 2) \\ n(n + 1)(n + 2) = \color{orange}n {}^{3} + \color{teal}3n {}^{2} + \color{grey}2n \\ \\ \underline{formula \: \: pt. \: \: n {}^{3} } : \\ 1 {}^{3} + 2 {}^{3} + 3 {}^{3} + ... + n {}^{3} =\boxed{ \color{orange} [(\tfrac{n(n + 1)}{2} ) ] {}^{ {}^{2} } } \\ \\ \underline{formula \: \: pt. \: \: 3n {}^{2} } : \\ 3(1 {}^{2} + 2 {}^{2} + 3 {}^{2} + ... + n {}^{2} ) \\ = 3 \cdot \tfrac{n(n + 1)(2n + 1)}{6} \\ = \boxed{ \color{teal}\tfrac{n(n + 1)(2n + 1)}{2} } \\ \\ \underline{formula \: \: pt. \: 2n} : \\ 2(1 + 2 + 3 + ... + n) \\ = 2 \cdot \tfrac{n \cdot(n + 1)}{2} \\ = \boxed{\color{grey}n(n + 1)} \\ \\ \\ \\ \\ \\ \implies \:\color{red} S\color{w}=\color{orange} [( \tfrac{n(n + 1)}{2} ) ] {}^{ {}^{2} } + \color{teal}\tfrac{n(n + 1)(2n + 1)}{2} +\color{grey} n(n + 1) \\ \color{white}=( \tfrac{n {}^{2} + n }{2} ) {}^{2} + \tfrac{(n {}^{2} + 2 )(2n + 1)}{2} + n {}^{2} + n \\ = \tfrac{n {}^{4} + 2n {}^{3} + n {}^{2} + 2(2n {}^{3} + 3n {}^{2} + n) + 4n {}^{2} + 4n}{4} \\ = \color{red}\boxed{ \tfrac{n {}^{4} + 6n {}^{3} + 11n {}^{2} + 6n}{4} }$