# Summation Formula

In mathematics, the summation is the basic addition of a sequence of any kind of 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 type of mathematical object can be associated with an operation called addition/summation which is denoted as “+”.

The summation of an explicit sequence is denoted as a succession of additions. For example, summation of (1, 3, 4, 7) can base denoted 1 + 3 + 4 + 7, and result for above notation is 15, that is, 1 + 3 + 4 + 7 = 15. Because 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?

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

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

### Where to use Summation?

Summation notation can be used in various field of mathematics like:

• 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=1âˆ‘n c = c + c + c + …. + c (n) times = nc

For example: Find the value of  i=1âˆ‘4 c.

By using property 1 we can directly calculate the value of i=1âˆ‘4 c as 4Ã—c = 4c.

Property 2

c=1âˆ‘n kc = (kÃ—1) + (kÃ—2) +  (kÃ—3)  + …. + (kÃ—n) …. (n) times = k Ã— (1 + … + n)  = k  c=1âˆ‘n c

For example: Find the value of i=1âˆ‘4 5i.

By using property 2 and 1 we can directly calculate value of i=1âˆ‘4 5i as 5 Ã— i=1âˆ‘4 i = 5 Ã— ( 1 + 2 + 3 + 4) = 50.

Property 3

c=1âˆ‘n (k+c) = (k+1) + (k+2) +  (k+3)  + …. + (k+n) …. (n) times = (n Ã— k) +  (1 + … + n)  = nk + c=1âˆ‘n c

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

By using property 2 and 3 we can directly calculate value of i=1âˆ‘4 (5+i) as 5Ã—4 +  i=1âˆ‘4 i = 20 + ( 1 + 2 + 3 + 4) = 30.

Property 4

k=1âˆ‘n (f(k) + g(k)) = k=1âˆ‘n f(k) + k=1âˆ‘n g(k)

For example: Find value of  i=1âˆ‘4 (i + i2).

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

### Some Standard Formula using the Summation

• Sum of first n natural numbers : (1+2+3+…+n) = i=1âˆ‘n (i) = [n Ã—(n +1)]/2
• Sum of square of first n natural numbers : (12+22+32+…+n2) = i=1âˆ‘n (i2) = [n Ã—(n +1)Ã— (2n+1)]/6
• Sum of cube of first n natural numbers : (13+23+33+…+n3) = i=1âˆ‘n (i3) = [n2 Ã—(n +1)2)]/4
• Sum of first n even natural numbers : (2+4+…+2n) = i=1âˆ‘n (2i) = [n Ã—(n +1)]
• Sum of first n odd natural numbers : (1+3+…+2n-1) = i=1âˆ‘n (2i-1) = n2
• Sum of square of first n even natural numbers : (22+42+…+(2n)2) = i=1âˆ‘n (2i)2 = [2n(n + 1)(2n + 1)] / 3
• Sum of square of first n odd natural numbers : (12+32+…+(2n-1)2) = i=1âˆ‘n (2i-1)2 = [n(2n+1)(2n-1)] / 3
• Sum of cube of first n even natural numbers : (23+43+…+(2n)3) = i=1âˆ‘n (2i)3 = 2[n(n+1)]2
• Sum of cube of first n odd natural numbers : (13+33+…+(2n-1)3) = i=1âˆ‘n (2i-1)3 = n2(2n2 – 1)

### Sample Problems

Problem 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=1âˆ‘n (i) = [n Ã—(n +1)]/2

We have sum of first 10 natural numbers =  i=1âˆ‘10 (i) = [10 Ã—(10 +1)]/2 = 55

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

Solution:

According to the question:

The sum of 10 first natural numbers greater than 5 =  i=6âˆ‘15 (i)

=  i=1âˆ‘15 (i) –  i=1âˆ‘5 (i)

= [15 Ã— 16 ] / 2 – [5 Ã— 6]/2

= 120 – 30

= 90

Problem 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=1âˆ‘8 i2 using the property/ formula of summation

i=1âˆ‘8 i2 = [8 Ã—(8 +1)Ã— (2Ã—8 +1)]/6 = [8 Ã— 9 Ã— 17] / 6

= 204

Problem 4: Simplify c=1âˆ‘n kc.

Solution:

Given summation formula = c=1âˆ‘n kc

= (kÃ—1) + (kÃ—2) + …… + (kÃ—n) (n terms)

= k (1 + 2 + 3 +….. + n)

c=1âˆ‘n kc = k c=1âˆ‘n c

Problem 5: Simplify and evaluate x=1âˆ‘n (4+x).

Solution:

Given summation is  x=1âˆ‘n (4+x)

As we know that c=1âˆ‘n (k+c) = nk + c=1âˆ‘n c

Given summation can be simplified as,

4n + x=1âˆ‘n (x)

Problem 6: Simplify x=1âˆ‘n (2x+x2).

Solution:

Given summation is  x=1âˆ‘n (2x+x2).

as we know that k=1âˆ‘n (f(k) + g(k)) = k=1âˆ‘n f(k) + k=1âˆ‘n g(k)

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

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