Summation Formula

In mathematics, the summation is the basic addition of a sequence of any numbers, called addends or summands; the result is their sum or total. In Mathematics numbers, functions, vectors, matrices, polynomials, and, in general, elements of any mathematical object can be associated with an operation called addition/summation, denoted as “+”.

Summation of an explicit sequence is denoted as a succession of additions. For example, the summation of (1, 3, 4, 7) can base denoted 1 + 3 + 4 + 7, and the result for the above notation is 15, that is, 1 + 3 + 4 + 7 = 15. Because the addition operation is associative as well as commutative, there is no need for parentheses while listing down the series/sequence, and the result is going to be the same irrespective of the order of the summands.

What is Summation Formula?

Summation or sigma (∑) notation is a method used to write out a long sum in a concise way. This notation can be attached to any formula or function.

For example, _i=1∑¹⁰ (i) is a sigma notation of the addition of finite sequence 1 + 2 + 3 + 4…… + 10 where the first element is 1 and the last element is 10.

Where to use the Summation Formula?

Summation notation can be used in various fields of mathematics:

• Sequence in series
• Integration
• Probability
• Permutation and Combination
• Statistics 

Note: A summation is a short form of repetitive addition. We can also replace summation with a loop of addition.

Properties of Summation

Property 1

_i=1∑ⁿ c = c + c + c + …. + c (n) times = nc 

For example: Find the value of  _i=1∑⁴ c.

By using property 1 we can directly calculate the value of _i=1∑⁴ c as 4×c = 4c.

Property 2

_c=1∑ⁿ kc = (k×1) + (k×2) +  (k×3)  + …. + (k×n) …. (n) times = k × (1 + … + n)  = k  _c=1∑ⁿ c

For example: Find the value of _i=1∑⁴ 5i.

By using property 2 and 1 we can directly calculate value of _i=1∑⁴ 5i as 5 × _i=1∑⁴ i = 5 × ( 1 + 2 + 3 + 4) = 50.

Property 3

_c=1∑ⁿ (k+c) = (k+1) + (k+2) +  (k+3)  + …. + (k+n) …. (n) times = (n × k) +  (1 + … + n)  = nk + _c=1∑ⁿ c

For example: Find the value of  _i=1∑⁴ (5+i).

By using property 2 and 3 we can directly calculate value of _i=1∑⁴ (5+i) as 5×4 +  _i=1∑⁴ i = 20 + ( 1 + 2 + 3 + 4) = 30.

Property 4

_k=1∑ⁿ (f(k) + g(k)) = _k=1∑ⁿ f(k) + _k=1∑ⁿ g(k)

For example: Find value of  _i=1∑⁴ (i + i²).

By using property 4  we can directly calculate value of _i=1∑⁴ (i + i²) as _i=1∑⁴ i  + _i=1∑⁴ i²  = ( 1 + 2 + 3 + 4) + (1 + 4 + 9 + 16) = 40.

Standard Summation Formulas

Various Summation formulas are,

Sum of First n Natural Numbers : (1+2+3+…+n) = _i=1∑ⁿ (i) = [n ×(n +1)]/2

Sum of Square of First n Natural Numbers : (1²+2²+3²+…+n²) = _i=1∑ⁿ (i²) = [n ×(n +1)× (2n+1)]/6

Sum of Cube of First n Natural Numbers : (1³+2³+3³+…+n³) = _i=1∑ⁿ (i³) = [n² ×(n +1)²)]/4

Sum of First n Even Natural Numbers : (2+4+…+2n) = _i=1∑ⁿ (2i) = [n ×(n +1)]

Sum of first n odd natural numbers : (1+3+…+2n-1) = _i=1∑ⁿ (2i-1) = n²

Sum of Square of First n Even Natural Numbers : (2²+4²+…+(2n)²) = _i=1∑ⁿ (2i)² = [2n(n + 1)(2n + 1)] / 3

Sum of Square of First n Odd Natural Numbers : (1²+3²+…+(2n-1)²) = _i=1∑ⁿ (2i-1)² = [n(2n+1)(2n-1)] / 3

Sum of Cube of First n Even Natural Numbers : (2³+4³+…+(2n)3) = _i=1∑ⁿ (2i)³ = 2[n(n+1)]²

Sum of Cube of First n Odd Natural Numbers : (1³+3³+…+(2n-1)³) = _i=1∑ⁿ (2i-1)³ = n²(2n² – 1)

Example on Summation Formula

Example 1: Find the sum of first 10 natural numbers, using the summation formula.

Solution: 

Using the summation formula for sum of n natural number _i=1∑ⁿ (i) = [n ×(n +1)]/2

We have sum of first 10 natural numbers =  _i=1∑¹⁰ (i) = [10 ×(10 +1)]/2 = 55

Example 2: Find the sum of 10 first natural numbers greater than 5, using the summation formula.

Solution: 

According to the question:

Sum of 10 first natural numbers greater than 5 =  _i=6∑¹⁵ (i) 

=  _i=1∑¹⁵ (i) –  _i=1∑⁵ (i) 

= [15 × 16 ] / 2 – [5 × 6]/2

= 120 – 15

= 105

Example 3: Find the sum of given finite sequence 1² + 2² + 3² + … 8².

Solution: 

Given sequence is 1² + 2² + 3² + … 8² , it can be written as  _i=1∑⁸ i² using the property/ formula of summation

_i=1∑⁸ i² = [8 ×(8 +1)× (2×8 +1)]/6 = [8 × 9 × 17] / 6 

= 204

Example 4: Simplify _c=1∑ⁿ kc.

Solution: 

Given summation formula = _c=1∑ⁿ kc

= (k×1) + (k×2) + …… + (k×n) (n terms)

= k (1 + 2 + 3 +….. + n)

_c=1∑ⁿ kc = k _c=1∑ⁿ c

Example 5: Simplify and evaluate x₌₁∑ⁿ (4+x).

Solution: 

Given summation is  _x=1∑ⁿ (4+x)

As we know that _c=1∑ⁿ (k+c) = nk + _c=1∑ⁿ c

Given summation can be simplified as,

4n + _x=1∑ⁿ (x)

Example 6: Simplify _x=1∑ⁿ (2x+x²).

Solution: 

Given summation is  _x=1∑ⁿ (2x+x²).

as we know that _k=1∑ⁿ (f(k) + g(k)) = _k=1∑ⁿ f(k) + _k=1∑ⁿ g(k)

given summation can be simplified as _x=1∑ⁿ (2x) + _x=1∑ⁿ (x²).