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Sum of Products of Consecutive Natural Numbers
## Formula
$$1\cdot2+2\cdot3+\cdots+n(n+1)=\frac{n(n+1)(n+2)}3$$
Therefore:
$$\sum_{k=1}^{n}k(k+1)=\frac{n(n+1)(n+2)}3$$
## Usage
- Used in summations involving products of consecutive integers.
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Sum of Products of Consecutive Natural Numbers
## Formula
$$1\cdot2+2\cdot3+\cdots+n(n+1)=\frac{n(n+1)(n+2)}3$$
Therefore:
$$\sum_{k=1}^{n}k(k+1)=\frac{n(n+1)(n+2)}3$$
## Usage
- Used in summations involving products of consecutive integers.
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