# How to calculate the problems of summation? Explained with examples

The summation is a wide concept in mathematics and statistics. It is frequently used in central tendencies, sum of squares, variance, standard deviation, algebraic sum of squares, and many other fields of education even in daily basis dealing.

The sum of different items or the sum of prices of different items are to be evaluated with the help of summation by simple summation or by using a sigma notation. In this lesson, we will learn all the basics of the summation along with solved examples.

## What is the summation?

In mathematics and statistics, the summation is a general notation for denoting the sum of the group of numbers or by making the list of numbers in a function by taking the initial and final terms on the Greek notation known as sigma notation.

The problems of the variance, standard deviation, arithmetic mean, geometric mean, and many other problems are solved easily with the help of summation notation as it is a compulsory part for getting the accurate result of the given problems.

In general, it is represented as:

Ui = U1 + U2 + U3 + … + Un

• = sigma notation or a notation of summation.
• Ui= given function.
• i = 1 is the starting value of the function.
• n = the last term of the function.

## Rules of summation notation

Here are the basic rules of summation notation that are used while solving the problems of sigma notation.

 Rule Name Rule Constant Rule K = n * k Sum Rule [f(t) + g(t)] = [f(t)] + [g(t)] Difference Rule [f(t) – g(t)] = [f(t)] – [g(t)] Product Rule [f(t) * g(t)] = [f(t)] * [g(t)] Quotient Rule [f(t) / g(t)] = [f(t)] / [g(t)]

## How to calculate the summation problems?

The summation of the given function is to be evaluated in two ways one is by substituting the values from initial to final in the whole function while the other way is to break the function by using laws of summation.

To ease up the calculations, you can get help form a summation calculator. This tool will give you the solution with steps to the given function in a fraction of a second.

Let us take an example to learn how to calculate the problems of summation in both ways.

Example

Evaluate the sum of the given function, if the initial value of the function is 3 and final value is 10.

f(t) = 4t3 + 2t2 – 49

Solution

First method:

Step 1:First of all, take the given information of the function and apply the notation of summation according to the given values.

f(t) = 4t3 + 2t2 – 49

initial value = 3

final value = 10

f(t) = [4t3 + 2t2 – 49]

Step 2:Now place the value to the given expression from initial to the final.

 For Substitute value of n to the function Simplification Result n = 3 4(3)3 + 2(3)2 – 49 4(27) + 2 (9) – 49 108 + 18 – 49 77 n = 4 4(4)3 + 2(4)2 – 49 4(64) + 2(16) – 49 256 + 32 – 49 239 n = 5 4(5)3 + 2(5)2 – 49 4(125) + 2(25) – 49 500 + 50 – 49 501 n = 6 4(6)3 + 2(6)2 – 49 4(216) + 2(36) – 49 864 + 72 – 49 887 n = 7 4(7)3 + 2(7)2 – 49 4(343) + 2(49) – 49 1372 + 98 – 49 1421 n = 8 4(8)3 + 2(8)2 – 49 4(512) + 2(64) – 49 2048 + 128 – 49 2127 n = 9 4(9)3 + 2(9)2 – 49 4(729) + 2(81) – 49 2916 + 162 – 49 3029 n = 10 4(10)3 + 2(10)2 – 49 4(1000) + 2(100) – 49 4000 +200 – 49 4151

Step 3: Now add all the results of n value from 3 to 10 to get the summation result.

[4t3 + 2t2 – 49] = 77 + 239 + 501 + 887 + 1421 + 2127 + 3029 + 4151

[4t3 + 2t2 – 49] = 12432

Second method:

Step 1:First of all, apply the notation of summation to each term of the function with the help of the sum and difference laws of sigma notation.

[4t3 + 2t2 – 49] = [4t3] + [2t2] – 

Now calculate each term of the function separately.

Step 2:For [4t3]

 For Substitute value of n to the function Simplification Result n = 3 4(3)3 = 4(27) = 108 n = 4 4(4)3 4(64) 256 n = 5 4(5)3 4(125) 500 n = 6 4(6)3 4(216) 864 n = 7 4(7)3 4(343) 1372 n = 8 4(8)3 4(512) 2048 n = 9 4(9)3 4(729) 2916 n = 10 4(10)3 4(1000) 4000

[4t3]= 108 + 256 + 500 + 864 + 1372 + 2048 + 2916 + 4000

[4t3]= 12064

Step 3: For [2t2]

 For Substitute value of n to the function Simplification Result n = 3 2(3)2 2 (9) 18 n = 4 2(4)2 2(16) 32 n = 5 2(5)2 2(25) 50 n = 6 2(6)2 2(36) 72 n = 7 2(7)2 2(49) 98 n = 8 2(8)2 2(64) 128 n = 9 2(9)2 2(81) 162 n = 10 2(10)2 2(100) 200

[2t2] = 18 + 32 + 50 + 72 + 98 + 128 + 162 + 200

[2t2] = 760

Step 4: For 

As 49 is a constant term so it will be n times the constant term.

 = 49 + 49 + 49 + 49 + 49 + 49 + 49 + 49

 = 49 x 8

 = 392

Step 5:Write the results of the summation functions and find the order of operation of them.

[4t3 + 2t2 – 49] = [4t3] + [2t2] – 

[4t3 + 2t2 – 49] = 12064 + 760 – 392

[4t3 + 2t2 – 49] = 12824 – 392

[4t3 + 2t2 – 49] = 12432

## Final words

Now you can grab all the basics of summation from this lesson as we have discussed all the basic intent of the topic with its definition and solved examples. You can solve the problems of summation easily by any method form the above discussed two methods.