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Sum of Squares of First $$n$$ Natural Numbers
## Formula
$$1^2+2^2+3^2+\cdots+n^2=\frac{n(n+1)(2n+1)}6$$
Therefore:
$$\sum_{k=1}^{n}k^2=\frac{n(n+1)(2n+1)}6$$
## Usage
- Used in summation problems involving squares of consecutive natural numbers.
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Sum of Squares of First $$n$$ Natural Numbers
## Formula
$$1^2+2^2+3^2+\cdots+n^2=\frac{n(n+1)(2n+1)}6$$
Therefore:
$$\sum_{k=1}^{n}k^2=\frac{n(n+1)(2n+1)}6$$
## Usage
- Used in summation problems involving squares of consecutive natural numbers.
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