Sum of squares of first n natural,even,odd numbers

Proofs are here.
The formula for the sum of squares of the first $n$ natural numbers is:

$$\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$$

Example:

For $n = 5$:

$$\sum_{k=1}^{5} k^2 = 1^2 + 2^2 + 3^2 + 4^2 + 5^2 = 55$$ $$\text{Using formula: }\frac{5(5+1)(2\cdot 5+1)}{6}=\frac{5\cdot 6\cdot 11}{6}=55$$

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The sum of squares of the first $n$ even natural numbers is given by:

$$\sum_{k=1}^{n} (2k)^2 = 4 \sum_{k=1}^{n} k^2 = 4 \cdot \frac{n(n+1)(2n+1)}{6}$$

So the final formula is:

$$\boxed{\frac{2n(n+1)(2n+1)}{3}}$$

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Example:

For $n = 3$ (i.e., even numbers: 2, 4, 6):

$$2^2 + 4^2 + 6^2 = 4 + 16 + 36 = 56$$ $$\text{Using formula: }\frac{2\cdot 3\cdot 4\cdot 7}{3}=\frac{168}{3}=56$$____________________________________________________________________

The sum of squares of the first $n$ odd natural numbers is given by:

$$\sum_{k=1}^{n} (2k - 1)^2 = \frac{n(2n - 1)(2n + 1)}{3}$$

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Example:

For $n = 3$ (i.e., odd numbers: 1, 3, 5):

$$1^2 + 3^2 + 5^2 = 1 + 9 + 25 = 35$$ $$\text{Using formula: } \frac{3(5)(7)}{3} = 35$$

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So the boxed formula is:

$$\boxed{{\displaystyle \frac{n(2n-1)(2n+1)}{3}}}$$

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The trick to remember them:
Remember the first master formula = n.(n+1).(2n+1)/6

Even sum is simply by multiplying it by 4: 2n.(n+1).(2n+1)/3

Odd sum is by moving the middle 1 in the previous formula to left.
So in:

2n.(n+1).(2n+1)/3

move +1 to left and make it -1:
(2n -1).n.(2n+1)/3