The **sum of squares** of the first \(n\) natural numbers refers to the following special sum,

\(\sum_{i=1}^n i^2 = \sum n^2 = 1^2 + 2^2 + 3^2 + \ldots + n^2\)

Up to the natural number \(n\).

Index

## Sum of Squares of n Natural Numbers Formula

We have the **formula for the sum of squares of n natural numbers** as follows:

\(\sum_{i=1}^n i^2 = \sum n^2 = \frac{n(n+1)(2n+1)}{6}\)

This gives us the sum of squares of consecutive natural numbers \(1^2 + 2^2 + 3^2 + \ldots + n^2\) upto \(n\), *for any natural number* \(n\).

## Variations of the Formula

We can have the following various sums based on the sum of squares of the first \(n\) natural numbers.

### Sum of the Squares of First n Even Natural Numbers

This sum is simply \(2^2 + 4^2 + 6^2 + \ldots + (2n)^2\) upto the \(n^{\text{th}}\) even natural number.

We can arrive at the formula as follows:

\(\begin{align}

\sum_{i=1}^n {(2i)}^2 &= 2^2 + 4^2 + 6^2 + \ldots + (2n)^2 \\

& = 2^2 (1^2) + 2^2 (2^2) + 2^2(3^2) + \ldots + 2^2 (n^2) \\

& = 2^2 (1^2 + 2^2 + 3^2 + \ldots + n^2)

\end{align}

\)

But we know that \(1^2 + 2^2 + 3^2 + \ldots + n^2\) is simply the sum of the squares of the first n natural numbers, which we know is \(\frac{n(n+1)(2n+1)}{6}\). Substituting this above, we get,

\(\begin{align}

\sum_{i=1}^n {(2i)}^2 &= 2^2 (1^2 + 2^2 + 3^2 + \ldots + n^2) \\

& = 4 (\frac{n(n+1)(2n+1)}{6}) \\

& = \frac{2n(n+1)(2n+1)}{3}

\end{align}

\)

Thus, we have** formula for sum of squares of first \(n\) even natural numbers**,

\(\sum_{i=1}^n {(2i)}^2 =2^2 + 4^2 + 6^2 + \ldots +(2n)^2= \frac{2n(n+1)(2n+1)}{3}\)

Read More Sum of Even Numbers Formula

### Sum of Squares of First n Odd Natural Numbers

This sum is simply written as \(1^2 + 3^2 + 5^2 + \ldots + (2n-1)^2\).

This can also be simply written as \(\sum_{i=1}^n (2i-1)^2\) or \(\sum {(2n-1)}^2\).

We can derive the formula by noting the following, **for the sums of squares of the first \(2n\) natural numbers. **

\(\sum_{i=1}^{2n} i^2 = 1^2 + 2^2 + 3^2 + \ldots + (2n)^2 = (1^2 + 3^2 + \ldots + (2n-1)^2) + (2^2 + 4^2 + \ldots + (2n)^2)\)

We simply rearranged the terms, to *group odd and even natural numbers together*. The sums on right are the sums of the squares of the first \(n\) odd and even numbers, respectively. Now we can write,

\(\sum_{i=1}^{2n} i^2 = 1^2 + 2^2 + 3^2 + \ldots + (2n)^2 = \sum_{i=1}^n (2i-1)^2 + \sum_{i=1}^n {(2i)}^2\)

This gives us, for the sum of the first \(n\) odd numbers,

\(\sum_{i=1}^n (2i-1)^2 = \sum_{i=1}^{2n} i^2 – \sum_{i=1}^n {(2i)}^2\)

We know the formula for the sum of the squares of the first \(n\) numbers, *so we just have to replace* \(n\) with \(2n\) for the **first part** of RHS.

We get, after replacing,

\(\begin{align}

\sum_{i=1}^n (2i-1)^2 &= \frac{2n(2n+1)(4n+1)}{6} – \frac{2n(n+1)(2n+1)}{3} \\

& = \frac{n(2n+1)}{3} [(4n+1) – 2(n+1)] \\

& = \frac{n(2n+1)}{3} (2n -1) \\

& =\frac{(n(2n+1)(2n-1)}{3}

\end{align}

\)

Thus we have **formula for sum of squares of first \(n\) odd numbers,**

\(\sum_{i=1}^n (2i-1)^2 = \sum (2n-1)^2 = \frac{(n(2n+1)(2n-1)}{3}\)

Read More Sum of Odd Number Formula

### Sum of Squares of Two Real Numbers

While this is not a direct application, it is still useful to know. Let \(x_1\) and \(x_2\) be **the two real numbers** whose squares we wish to add. We have,

\(x_1^2 + x_2^2 = (x_1 + x_2)^2 – 2{x_1}{x_2}\)

### Sum of Squares of Three Real Numbers

Let the **three real numbers be** \(x_1\), \(x_2\) and \(x_3\). We have, from elementary identities,

\(x_1^2 + x_2^2 + x_3^2 = (x_1+x_2+x_3)^2 – 2[(x_1x_2)+(x_2x_3)+(x_3x_1)]\)

## Solved Examples

**Question 1.** Find the sum of the squares of the first 20 natural numbers.

**Solution.** We know that the formula for the sum of the squares of the first n natural numbers, the formula is,

\(\sum_{i=1}^n i^2 = \sum n^2 = \frac{n(n+1)(2n+1)}{6}\)

Substituting \(n=20\) for the above equation, we get,

\(\sum_{i=1}^{20} i^2 = \frac{20(20+1)(2(20)+1)}{6}\)

Upon simplification, we get,

\(\sum_{i=1}^{20} i^2 = 2870\)

**Question 2.** Find the sum of the squares of the first 18 odd numbers.

**Solution.** We know that the sum of the squares of the first n odd numbers is,

\(\sum_{i=1}^n (2i-1)^2 = \sum (2n-1)^2 = \frac{(n(2n+1)(2n-1)}{3}\)

Substituting \(n=18\) in the above equation, we have,

\(\sum_{i=1}^{18} (2i-1)^2 = \frac{(18(2(18)+1)(2(18)-1)}{3}\)

Calculating, we have,

\(\sum_{i=1}^{18} (2i-1)^2 = 7770\)

## FAQs

**How to find sum of squares of n natural numbers?**We can find the sum of the squares of n natural numbers by a formula,

\(\sum_{i=1}^n i^2 = \sum n^2 = \frac{n(n+1)(2n+1)}{6}\)

**What is the sum of the squares of the first 100 natural numbers?**The sum of the squares of the first 100 natural numbers can be obtained by substituting n=100 in the above formula.

We get the sum of the squares of the first 100 natural numbers as 338350.

**What is difference in formula for sum of squares of first n natural numbers and sum of first n numbers?**The formula for sum of first \(n\) natural numbers is,

\(\sum n = \frac{n(n+1)}{2}\)*On the other hand*, the sum of the squares of first \(n\) natural numbers is:

\(\sum n^2 = \frac{n(n+1)(2n+1)}{6}\)

**What is the sum of the first n odd numbers?**The formula for the sum of the first \(n\) odd numbers is simply \(n^2\).