∑ i = 1 n i = n ( n + 1 ) 2 {\displaystyle \sum _{i=1}^{n}i={\frac {n(n+1)}{2}}} ∑ i = 1 n i ( i + 1 ) = n ( n + 1 ) ( n + 2 ) 3 {\displaystyle \sum _{i=1}^{n}i(i+1)={\frac {n(n+1)(n+2)}{3}}} ∑ i = 1 n i ( i + 1 ) ( i + 2 ) = n ( n + 1 ) ( n + 2 ) ( n + 3 ) 4 {\displaystyle \sum _{i=1}^{n}i(i+1)(i+2)={\frac {n(n+1)(n+2)(n+3)}{4}}} ∑ i = 1 n i ( i + 1 ) ( i + 2 ) ( i + 3 ) = n ( n + 1 ) ( n + 2 ) ( n + 3 ) ( n + 4 ) 5 {\displaystyle \sum _{i=1}^{n}i(i+1)(i+2)(i+3)={\frac {n(n+1)(n+2)(n+3)(n+4)}{5}}} 請問如何證明這個規律會持續下去？謝謝！
