To solve the problem of finding the sum ( S = 1 \times 2 + 2 \times 3 + 3 \times 4 + \dots + n \times (n+1) ), follow these steps:

Step 1: Expand each term

Each term ( k \times (k+1) = k^2 + k ). Thus, the sum becomes:
[ S = \sum{k=1}^n (k^2 + k) = \sum{k=1}^n k^2 + \sum_{k=1}^n k ]

Step 2: Use known summation formulas

• Sum of squares: ( \sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6} )
• Sum of first ( n ) integers: ( \sum_{k=1}^n k = \frac{n(n+1)}{2} )

Step 3: Combine the formulas

[ S = \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} ]

Factor out ( \frac{n(n+1)}{6} ):
[ S = \frac{n(n+1)}{6} \left[ (2n+1) + 3 \right] = \frac{n(n+1)(2n+4)}{6} ]

Simplify:
[ S = \frac{n(n+1) \times 2(n+2)}{6} = \frac{n(n+1)(n+2)}{3} ]

Answer: (\boxed{\dfrac{n(n+1)(n+2)}{3}})

作者声明:本文包含人工智能生成内容。