# Sum of Squares – Definition, Formula, Examples, FAQs

**Sum of squares** refers to the sum of the squares of the given numbers, i.e., it is the addition of squared numbers. The squared terms could be two terms, three terms, or “n” number of terms, the first “n” odd or even terms, a series of natural numbers or consecutive numbers, etc. In statistics, the value of the sum of squares tells the degree of dispersion in a dataset. For this, we need to first find the mean of the given data, then the variation of each data point from the mean, square them, and finally, add them. In algebra, we use the (a + b)^{2} identity to determine the sum of the squares of two numbers. The formula that determines the sum of the squares of the first “n” natural numbers is derived with the help of the sum of the squares of the first “n” natural numbers. We perform these fundamental arithmetic operations, which are necessary for both algebra and statistics. There are various methods to determine the sum of squares of given numbers.

## Sum of Squares Formula

The sum of the square formula is appliable for two, three, and up to n terms which are explained below:

### Sum of squares for two numbers

Let a and b be two real numbers, then the formula for the addition of squares of the two numbers is given as follows:

a^{2}+ b^{2}= (a + b)^{2}− 2ab

**Proof:**

From the algebraic identities, we have,

(a + b)

^{2}= a^{2}+ 2ab + b^{2}Now, subtract 2ab on both sides.

(a + b)

^{2}− 2ab = a^{2}+ 2ab + b^{2}− 2aba

^{2}+ b^{2}= (a + b)^{2}− 2abHence, proved.

### Sum of squares for three numbers

Let a, b, and c be three real numbers, then the formula for the addition of squares of the three numbers is given as follows:

a^{2}+ b^{2}+ c^{2}= (a +b + c)^{2}− 2ab − 2bc − 2ca

**Proof:**

From the algebraic identities, we have,

(a + b + c)

^{2}= a^{2}+ b^{2}+ c^{2}+ 2ab + 2bc + 2caBy subtracting 2ab, 2bc, and 2ca on both sides, we get,

a

^{2}+ b^{2}+ c^{2 }= (a + b + c)^{2}− 2ab − 2bc − 2caHence, proved

## Sum of squares for “n” Natural Numbers

Natural numbers are also known as positive integers and include all the counting numbers, starting from 1 to infinity. If 1, 2, 3, 4,… n are n consecutive natural numbers, then the sum of squares of “n” consecutive natural numbers is represented by 1^{2} + 2^{2} + 3^{2} +… + n^{2} and symbolically represented as Σn^{2}.

The formula for the sum of squares of the first “n” natural numbers is given as follows:

∑n^{2}= [n(n+1)(2n+1)]/6

### Sum of Squares of First “n” Even Numbers

The formula for the sum of squares of the first “n” even numbers, i.e., 2^{2} + 4^{2} + 6^{2} +… + (2n)^{2} is given as follows:

∑(2n)^{2} = 2^{2 }+ 4^{2} + 6^{2} +… + (2n)^{2}

∑(2n)^{2}= [2n(n+1)(2n+1)]/3

### Sum of Squares of First “n” Odd Numbers

The formula for the sum of squares of the first “n” odd numbers, i.e., 1^{2} + 3^{2} + 5^{2} +… + (2n – 1)^{2}, can be derived using the formulas for the sum of the squares of the first “2n” natural numbers and the sum of squares of the first “n” even numbers.

∑(2n-1)^{2} = 1^{2} + 3^{2} + 5^{2} + … + (2n – 1)^{2}

∑(2n-1)^{2}= [n(2n+1)(2n-1)]/3

**Proof:**

∑(2n –1)

^{2}= [1^{2}+ 2^{2}+ 3^{2}+ … + (2n – 1)^{2}+ (2n)^{2}] – [2^{2}+ 4^{2}+ 6^{2}+ … + (2n)^{2}]Now, apply the formula for the addition of squares of “2n” natural numbers and “n” even natural numbers, and we get;

∑(2n–1)

^{2}= 2n/6 (2n + 1)(4n + 1) – (2n/3) (n+1)(2n+1)∑(2n–1)

^{2}= n/3 [(2n+1)(4n+1)] – 2n/3 [(n+1)(2n+1)]Now, take out the common terms.

∑(2n–1)

^{2}= n/3 (2n+1) [4n + 1 – 2n – 2]∑(2n–1)

^{2}= [n(2n+1)(2n–1)]/3Hence, proved.

## Sum of Squares in Statistics

Sum of squares of n data points = ∑^{n}_{i=0}(x_{i}– x̄)^{2}∑ = represents sum

x_{i}= each value in the set

x̄ = mean of the values

x_{i}– x̄ = deviation from the mean value

(x_{i }– x̄)^{2}= square of the deviation

n = number of terms in the series

In statistics, the value of the sum of squares tells the degree of dispersion in a dataset. It evaluates the variance of the data points from the mean and aids in a better understanding of the data. The large value of the sum of squares indicates that there is a high variation of the data points from the mean value, while the small value indicates that there is a low variation of the data from its mean. Follow the steps given below to find the total sum of squares in statistics.

**Step 1:**Count the number of data points in the given dataset.**Step 2:**Now, calculate the mean of the given data.**Step 3:**Subtract each data point from the mean calculated in step 2.**Step 4:**Now, determine the square of the difference obtained in step 3.**Step 5:**Finally, add the squares that we have determined in step 4.

## Solved Examples based on Sum of Squares

**Example 1: Find the sum of the given series: 1 ^{2 }+ 2^{2} + 3^{2} +…+ 55^{2}.**

**Solution:**

To find the value of 1

^{2}+ 2^{2}+ 3^{2}+…+ 55^{2}.From the sum of squares formula for n terms, we have

∑n

^{2}= 1^{2}+ 2^{2}+ 3^{2}+…+ n^{2}= [n(n+1)(2n+1)] / 6Given, n = 55

= [55(55+1)(2×55+1)] / 6

= (55 × 56 × 111) / 6

= 56,980

Thus, the sum of the given series is 56,980.

**Example 2: Find the value of (3 ^{2} + 8^{2}), using the sum of squares formula.**

**Solution:**

To find the value of 3

^{2}+ 8^{2}Given: a = 3 and b = 8.

From the sum of squares formula, we have

a

^{2}+ b^{2}= (a + b)^{2}− 2ab⇒ 3

^{2}+ 8^{2}= (3 + 8)^{2}− 2(3)(8)= 121 – 2(24)

= 121 − 48

= 73.

Hence, the value of (3

^{2}+ 8^{2}) is 73.

**Example 3: Find the sum of squares of the first 25 even natural numbers.**

**Solution:**

To find the value of 2

^{2 }+ 4^{2}+ 6^{2}+… + 48^{2}+ 50^{2}.= 2

^{2}( 1^{2}+ 2^{2}+ 3^{2}+…+25^{2})From the sum of squares formula for n terms, we have

∑n

^{2}= [n(n+1)(2n+1)]/6Here, n = 25

2

^{2}( 1^{2}+ 2^{2}+ 3^{2}+…+25^{2}) = 4[25(25+1)(2(25)+1)/6]= (2/3) × (25) × (26) × (51)

= 22,100

Hence, the sum of squares of the first 25 even natural numbers is 22,100.

**Example 4: ** **A dataset has points 2, 4, 13, 10, 12, and 7. Find the sum of squares for the given data.**

**Solution:**

Given: We have 6 data points 2, 4, 13, 10, 12, and 7.

The sum of the given data points = 2 + 4 + 13 + 10 + 12 + 7 = 48.

The mean of the given data is given by,

Mean, x̄ = Sum / Number of data points

= 48 / 6

= 8

So, the sum of squares is given by,

∑

^{n}_{i}=0 (x_{i}– x̄)^{2}= (2 – 8)^{2}+ (4 – 8)^{2}+ (13 – 8)^{2}+ (10 – 8)^{2}+ (12 – 8)^{2}+ (7 – 8)^{2}= (–6)

^{2}+ (–4)^{2}+ (5)^{2}+ (2)^{2}+ (4)^{2}+ (–1)^{2}= 36 + 16 + 25 + 4 + 14 + 1

= 96

Hence, the sum of squares for the given data is 96.

**Example 5: Calculate the sum of the squares of 4, 9, and 11 using the sum of squares formula for three numbers.**

**Solution: **

To find the value of 4, 9, and 11.

Given, a = 4, b = 9, and c = 11.

From the sum of squares formula, we have

a

^{2}+ b^{2}+ c^{2}= (a + b +c)^{2 }− 2ab − 2bc − 2ca4

^{2}+ 9^{2}+ 11^{2}= (4 + 9 + 11)^{2}−(2×4×9) − (2×9×11) − (2×11×4)= 576 − 72 − 198 − 88

= 218

Hence, the value of (4

^{2}+ 9^{2}+ 11^{2}) is 218.

**Example 6: Find the sum of squares of the first 10 odd numbers.**

**Solution:**

The sum of squares of the first 10 odd numbers: 1

^{2}+ 3^{2}+ 5^{2}+… +17^{2}+ 19^{2}We know that,

The sum of squares of first “n” Odd Numbers ∑(2n–1)

^{2}= [n(2n+1)(2n–1)]/3Here, n is 10.

= [10×(2×10 + 1)(2×10 – 1)]/3

= [10 × 21 × 19]/3

= 10 × 7 × 19 = 1,330

Hence, the value of the sum of squares of the first 10 odd numbers is 1330.

## FAQs based on Sum of Squares

**Question 1: What is the Sum of Squares Error?**

**Answer:**

Sum of squares error, also known as the residual sum of squares, is the difference between the actual value and the predicted value of the data.

**Question 2: What Is the Expansion of Sum of Squares Formula?**

**Answer:**

a

^{2}+ b^{2}formula is known as the sum of squares formula in algebra and it is read as a square plus b square. Its expansion is expressed as a2 + b2 = (a + b)2 – 2ab.

**Question 3: Write the Sum of Squares Formula used in Algebra.**

**Answer: **

The sum of squares formula used in algebra are:

a^{2}+ b^{2}= (a + b)^{2}− 2aba^{2}+ b^{2}+ c^{2}= (a +b + c)^{2}− 2ab − 2bc − 2ca

**Question 4: Write the sum of squares of the first five even numbers.**

**Answer:**

The sum of squares of the first five even numbers is given by:

∑(2n)^{2}= [2n(n+1)(2n+1)]/3putting n = 5

∑(2×5)

^{2}= [2×5×(5+1)×(2×5+1)]/3= 220

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