Sum of Squares: Formula, Derivation & Examples

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The sum of square numbers is known as the sum of squares. The sum of square denotes the square of two terms, three terms or n number of terms. The Sum of squares is a basic operation used in statistics, algebra and numbers series. In statistics, it is used to find the variation in the data. In Algebra and numbers series, it is used as a basic arithmetic operation.

Keyterms: Sum, Squares, Statistics, Algebra, Numbers series, Arithmetic operation, Numbers, Deviation

Read More: NCERT Solutions for Class 11 Math Sequences and Series


Sum of Squares in Statistics

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In statistics, the sum of squares is also known as a total sum of squares and it is noted by TSS. The formula of a total sum of squares is given below,

TSS = ∑(xi - x)

Where,

Xi → Value in the set of numbers

x‾ → Mean of Numbers

i → 1,2,3,……, n

xi - x‾ → Deviation of Values

(xi + x‾) → Square of the Deviation

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Sum of Squares in Algebra

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Algebraic formulas are useful to calculate the squares of large numbers easily.

Let a and b be the two natural numbers then the Formula to find the sum of squares of a and b is

(a+b)2 = a2 + b2 + 2ab

→ a2 + b2 = (a+b)2 - 2ab

The below tables illustrates the proof of the above formula.

a b
a a2 ab
b ab b2

Sum of three squares in algebra

Let a, b and c be the three natural numbers, the formula to find the sum of squares of a, b and c is given as follows.

 (a+b+c)2 = a2+b2+c2+2ab+2ac+2bc

→ a2+b2+c2 = (a+b+c)2-2ab-2ac-2bc

The below tables illustrates the proof of the above formula.

a b c
a a2 ab ac
b ab b2 bc
c ac bc c2

Sum of squares of natural numbers

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Natural numbers are numbers used for counting and ordering ranges from 1 to infinity and it is also known as positive integers. The formulas to calculate the sum of squares of n number of natural numbers, n even numbers and n odd numbers are tabulated below.

Sum of Squares Formula
Sum of Squares of n natural numbers \(\frac{ {n(n+1) (2n+1)} } {6}\)
Sum of Squares of n even numbers \(\frac { {2n(n+1) (2n+1)} } {3}\)
Sum of Squares of n odd numbers \(\frac { {n(n+1) (2n-1)} } {3}\)

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Formula Derivation of Sum of Squares

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The formula derivation of the sum of squares is given below.

Sum of squares of n natural numbers 

The formula is proved using the algebraic expression. The sum of the square of n natural numbers is denoted by ∑n2. The algebraic expression used to prove this formula is 

a3 - b3 = (a-b) (a2+ ab + b2

Replace a as n and b as (n-1)

→ n3 - (n-1)3 = (n - (n-1))(n2 + n(n-1) + (n-1)2)

→ n3 - (n-1)3 = 1(n+ n2 - n + n2 + 2n+1)

→ n3 - (n-1)3 = 3n2 + n + 1

By substituting n= 1, 2…, n successively, we get

13 – 03 = 3 (1)2 – 3 (1) + 1

23 – 13 = 3 (2)– 3 (2) + 1

33 – 23 = 3 (3)2 – 3 (3) + 1

Solving on both sides, we can get

→ n3 - 03 = 3∑n2 -3∑n + n

We know that, ∑n = (n(n+1))/2

→ ∑n2 = (1/3) {n+ (3n(n+1)/2) - n}

→ ∑n2 = (1/6)(2n+ 3n2 + n)

Therefore,

∑n2 = {n(n+1) (2n+1)}6

Sum of squares of n even numbers

The even number is denoted by 2n. The derivation of the sum of squares formula is derived below.

∑(2n)2 = 22+42+62+……..+(2n)2

→ ∑(2n)2 = 2(12+22+32+........+n2)

By applying a sum of squares of n natural number formula in the above equation We get,

→ ∑(2n)2 = 2(( n x (n+1) x (2n+1))/6)

Therefore,

∑(2n)2 ={2n (n+1) (2n+1)}3

Sum of squares of n odd numbers

The formula is derived by adding the sum of squares of 2n natural numbers and

n even numbers.

Σ(2n-1)2 = ({(2n/6)(2n+1)}(4n+1))-((2n/3)(n+1)(2n+1))

→ Σ(2n-1)2=({(n/3)(2n+1)}(4n+1))-2(n+1)

→ Σ(2n-1)2=({(n/3)(2n+1)}(4n+1-2n-1)

Therefore,

Σ(2n-1)2 = {n(2n+1) (2n-1)}3

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Things to Remember

  • The sum of squares formulas is used to find the sum of squares of large numbers in an easy way. It is used in statistics to find the variance of a given value. In algebra and number series it is used as a basic arithmetic operation.
  • The natural number is divided into two types, they are even numbers are odd numbers. The sum of two odd numbers is always an even number and the product of two or more odd numbers is always an odd number.
  • Always we have to use algebraic expressions to derive the formula for the sum of squares.
  • Sum of Squares of n natural numbers is given by {n(n+1) (2n+1)}6
  • Sum of Squares of n even numbers is given by {2n(n+1) (2n+1)}3
  • Sum of Squares of n odd numbers is given by {n(n+1) (2n-1)}3

Previous Years’ Questions (PYQs)

  1. If the 2nd, 5th and 9th terms of a non-constant… (JEE Main 2016)
  2. If 4= 16= 64z, then… (JKCET 2011)
  3. Sum of n terms of the series 1 + 11…
  4. Sum to 10 terms of the series 1+2…
  5. The 10th common term between the series
  6. The sum of squares of deviations for 10 observations taken from mean 50…
  7. If Sm denotes the sum of first m terms of a G.P. of n terms
  8. The 100th term of the sequence 1, 2, 2, 3, 3, 3,....
  9. A sequence containing a finite number of terms is called a…
  10. For a positive integer n let a(n)... (JEE Advanced 1999)
  11. If the value of mode and mean is 60 …? (JEEA 2013)
  12. The mean square deviation of a set of nn observation x1,x2,...xabout a point cc is defined as 1n∑i=1n(xi−c)2…? (BITS Admission Test 2015)
  13. The arithmetic mean of the data 0,1,2,……,n with frequencies…?(BITS Admission Test 2015)
  14. If coefficient of variation is 60 and standard deviation is 24, then Arithmetic mean is?  (KCET 2017)

Sample Questions

Ques. Calculate the sum of squares of 13 and 14 using algebraic identities and prove LHS and RHS are equal. (3 marks)

Ans. To calculate the sum of squares of two numbers, we can use the algebraic identities,

a2+b2 = (a+b)2-2ab

RHS: (a+b)2-2ab = (13+14)2-(2x13x14)

(a+b)2-2ab = 729-364

(a+b)2-2ab = 365

LHS: a2+b2 = 132+142

a2+b2 = 169+196

a2+b2 = 365

LHS = RHS

Hence Proved.

Ques. Calculate the sum of squares of 6, 7, 9 using algebraic identities and prove LHS and RHS are equal. (3 marks)

Ans. To calculate the sum of squares of three numbers, we can use the algebraic identities,

a2+b2+c2 = (a+b+c)2-2ab-2ac-2bc

Substitute a=6, b=7 and c=9

RHS: (a+b+c)2-2ab-2ac-2bc = (6+7+9)2-(2x6x7)-(2x6x9)-(2x7x9)

(a+b+c)2-2ab-2ac-2bc = (484)-(84)-(108)-(126)

(a+b+c)2-2ab-2ac-2bc = 166

LHS: a2+b2+c2 = 62+72+92

a2+b2+c2 = 36+49+81

a2+b2+c2 = 166

LHS = RHS

Hence Proved.

Ques. Calculate the sum of squares of first n natural numbers, where n=5 and prove LHS is equal to RHS. (3 marks)

Ans. To calculate the sum of squares of first n natural number is

∑n2= (n x (n+1) x (2n+1))/6

Given n=5,

RHS: (n x (n+1) x (2n+1))/6 = (5x6x11)/6

(n x (n+1) x (2n+1))/6 = 55

LHS: ∑n2 = 1+4+9+16+25 = 55

∑n2 = 55

LHS = RHS

Hence Proved

Ques. Calculate the sum of squares of first n even numbers, where n=4 and prove LHS is equal to RHS. (3 marks)

Ans. To calculate the sum of squares of first n even number is

∑(2n)=( n x (n+1) x (2n+1))/3

Given n=4,

RHS: ( n x (n+1) x (2n+1))/3 = (4x5x9)/3

( n x (n+1) x (2n+1))/3 = 60

LHS: ∑(2n)2 = 2(12+22+32+42)

∑(2n)2 = 2(1+4+9+16)

∑(2n)= 2x30

∑(2n)2 = 60

LHS = RHS

Hence Proved

Ques. Calculate the sum of squares of first n odd numbers, where n=7 and prove LHS is equal to RHS. (3 marks)

Ans. To calculate the sum of squares of first n even number is

Σ(2n-1)2=(n x (2n+1) x (2n-1))/3

Given n=7,

RHS: (n x (2n+1) x (2n-1))/3 = (7x15x13)/3

(n x (2n+1) x (2n-1))/3 = 455

LHS: Σ(2n-1)2 = 12+32+52+72+92+112+132

Σ(2n-1)2= 455

LHS = RHS

Hence Proved

Ques. Calculate the sum of squares of first 6 natural numbers. (3 marks)

Ans. Given n=6,

RHS, ( n x (n+1) x (2n+1))/3 = (6x7x13)/3

( n x (n+1) x (2n+1))/3 = 182

LHS, ∑(2n)2 = 22+42+62+82+102+122

∑(2n)2 = 2(1+4+9+16+25+36)

∑(2n)2 = 182

LHS=RHS

Hence proved

Ques. Check whether (6, 8, 10) is a Pythagorean triplet. (3 marks)

Ans. 2m, m2 – 1 and m2 + 1 represent the Pythagorean triplet.

Let 2m = 6 ⇒ m = 3

m2 – 1 = (3)2 – 1 = 9 – 1 = 8

and m2 + 1 = (3)2 + 1 = 9 + 1 = 10

Hence (6, 8, 10) is a Pythagorean triplet.

Alternative Method:

(6)2 + (8)2 = 36 + 64 = 100 = (10)2

⇒ (6, 8, 10) is a Pythagorean triplet.

Ques. Which of the following triplets is Pythagorean? (4 marks)
(i) (14, 48, 50)
(ii) (18, 79, 82)

Ans. We know that 2m, m2 – 1 and m2 + 1 make Pythagorean triplets.

(i) For (14, 48, 50),

Put 2m =14 ⇒ m = 7

m2 – 1 = (7)2 – 1 = 49 – 1 = 48

m2 + 1 = (7)2 + 1 = 49 + 1 = 50

Hence (14, 48, 50) is a Pythagorean triplet.

(ii) For (18, 79, 82)

Put 2m = 18 ⇒ m = 9

m2 – 1 = (9)2 – 1 = 81 – 1 = 80

m2 + 1 = (9)2 + 1 = 81 + 1 = 82

Hence (18, 79, 82) is not a Pythagorean triplet.

Ques. Find two consecutive numbers whose squares have the sum 85. (CBSE 2012) (3 marks)

Ans. Let first number = x

Then second number = x + 1

According to the condition

x2 + (x + 1)2 = 85

⇒ x2 + x+ 2x + 1 = 85

⇒ 2x2 + 2x + 1 – 85 = 0

⇒ 2x2 + 2x – 84 = 0

⇒ x2 + x – 42 = 0

⇒ x2 + 7x – 6x – 42 = 0

⇒ x (x + 7) – 6 (x + 7) = 0

⇒ (x + 7) (x – 6) = 0

Either x + 7 = 0, then x = -7 or x – 6 = 0, then x = 6

(i) If x = -7, then the first number = -7 and second number = -7 + 1 = -6

(ii) If x = 6, then the first number = 6 and second number = 6 + 1 = 7

Hence numbers are -7, -6 or 6, 7

Ques. Divide 29 into two parts so that the sum of the squares of the parts is 425. (3 marks)

Ans. Total = 29

Let first part = x

Then second part = 29 – x

According to the condition

x2 + (29 – x)2 = 425

⇒ x2 + 841 + x2 – 58x = 425

⇒ 2x2 – 58x + 841 – 425 = 0

⇒ 2x2 – 58x + 416 = 0

⇒ x2 – 29x + 208 = 0 (Dividing by 2)

⇒ x2 – 13x – 16x + 208 = 0

⇒ x(x – 13) – 16 (x – 13) = 0

⇒ (x – 13) (x – 16) = 0

Either x – 13 = 0, then x = 13 or x – 16 = 0, then x = 16

(i) If x = 13, then First part =13 and second part = 29 – 13 = 16

(ii) If x = 16, then First part =16 and second part = 29 – 16 = 13

Parts are 13, 16


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CBSE CLASS XII Related Questions

1.

Evaluate \(\begin{vmatrix} cos\alpha cos\beta &cos\alpha sin\beta  &-sin\alpha \\   -sin\beta&cos\beta  &0 \\   sin\alpha cos\beta&sin\alpha\sin\beta  &cos\alpha  \end{vmatrix}\)

      2.
      Let f: R→R be defined as f(x) = 3x. Choose the correct answer.

        • f is one-one onto
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        3.
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        (ii) \((A-B)'=A'-B'\)

            4.

            If A=\(\begin{bmatrix}2&-1&1\\-1&2&-1\\1&-1&2\end{bmatrix}\)verify that A3-6A2+9A-4 I=0 and hence find A-1 

                5.
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                    6.

                    Let A=\(\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}\),show that(aI+bA)n=anI+nan-1bA,where I is the identity matrix of order 2 and n∈N

                        CBSE CLASS XII Previous Year Papers

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