Summation Formulas

Before going to learn summation formulas, first, we will recall the meaning of summation. Summation (or) sum is the sum of consecutive terms of a sequence. To write the sum of more terms, say n terms, of a sequence \(\{a_n\}\), we use the summation notation instead of writing the whole sum manually. i.e., \(a_1+a_2+...+a_n= \sum_{i=1}^{n} a_{i}\). Let us learn the summation formulas and their applications using a few solved examples.

What Are Summation Formulas?

The summation formulas are used to calculate the sum of the sequence. There are various types of sequences such as arithmetic sequence, geometric sequence, etc and hence there are various types of summation formulas of different sequences. Also, there are summation formulas to find the sum of the natural numbers, the sum of squares of natural numbers, the sum of cubes of natural numbers, the sum of even numbers, the sum of odd numbers, etc. Here is the list of summation formulas. We will learn each of these formulas in detail in the upcoming section.

List of Summation Formulas

We know that the sum of two numbers is the result obtained by adding two numbers. Thus, if \(\{x_{1}, x_{2},…,x_{n}\}\) is a sequence, then the sum of its terms is denoted using the symbol Σ (sigma). i.e., the sum of the above sequence = \(\sum_{i=1}^{n}x_{i}=x_{1}+x_{2}+….x_{n}\). Here, \(\sum_{i=1}^{n}\) represents the sum of the terms of the sequence from the 1^st term to the n^th term and it is read as "sigma i is equal to 1 to n". But we actually do not need to add the sum of the sequences manually all the time to find the sum. Instead, we use the following summation formulas. Here are some popular summation formulas.

• The sum of first n natural numbers is calculated using the formula: 
  \(\sum_{i=1}^{n} i\) = 1 + 2 + 3 + ... + n = \(\dfrac{n(n+1)}{2}\)

• The sum of the squares of the first n natural numbers is calculated using the formula: 
  \(\sum_{i=1}^{n} i^{2}\) = 1² + 2² + 3² + ... + n² = \(\dfrac{n(n+1)(2 n+1)}{6}\)

• The sum of the cubes of the first n natural numbers is calculated using the formula: 
  \(\sum_{i=1}^{n} i^{3}\) = 1³ + 2³ + 3³ + ... + n³ = \(\dfrac{n^{2}(n+1)^{2}}{4}\)

• The sum of the fourth powers of the first n natural numbers is calculated using the formula: 
  \(\sum_{i=1}^{n} i^{4}\) = 1⁴ + 2⁴ + 3⁴ + ... + n⁴ = \(\dfrac{1}{30} n(n+1)(2 n+1)\left(3 n^{2}+3 n-1\right)\)

• The sum of the first n even natural numbers is:
  \(\sum_{i=1}^{n}\) 2 i = 2 + 4 + 6 + ... (n numbers) = n (n + 1)

• The sum of the first n odd natural numbers is:
  \(\sum_{i=1}^{n}\) (2 i +1) = 1 + 3 + 5 + .... (n numbers) = n²

• The sum of the arithmetic sequence a, a + d, a + 2d, ... , a + (n - 1) d is:
  \(\sum_{i=1}^{n} a+(i-1) d=\dfrac{n}{2}[2 a+(n-1) d]\)

• For the geometric sequence a, ar, ar², ... , ar^{n - 1}, 
  The sum of the first n terms is, \(\sum_{i=1}^{n} a r^{i-1}=\dfrac{a\left(1-r^{n}\right)}{1-r} \)
  The sum of the infinite terms is, \(\sum_{i=1}^{\infty} a r^{i-1}=\dfrac{a}{1-r}\) (only when |r| < 1)

We will see the applications of the summation formulas in the upcoming section.

Examples Using Summation Formulas

Example 1: Find the sum of all even numbers from 1 to 100.

Solution:

We know that the number of even numbers from 1 to 100 is n = 50.

Using the summation formulas, the sum of the first n even numbers is 

n (n + 1) = 50 (50 + 1) = 50 (51) = 2550

Answer: The required sum = 2,550.

Example 2: Find the value of \(\sum_{i=1}^{n} (3-2i)\) using the summation formulas.

Solution:

To find: The given sum using the summation formulas.

\( \begin{align} &\sum_{i=1}^{n} (3 -2i)\\[0.2cm]&= 3 \sum_{i=1}^{n} 1 - 2 \sum_{i=1}^{n} i\\[0.2cm] &= 3 n - 2 \left( \dfrac{n(n+1)}{2} \right)\\[0.2cm] &= \dfrac{6n -2n^2-2n}{2}\\[0.2cm] &= \dfrac{4n-2n^2}{2}\\[0.2cm] &= 2n-n^2 \end{align}\)

Answer:  \(\sum_{i=1}^{n} (3 -2i) =2n-n^2\).

Example 3: Find the value of the summation \(\sum_{k=1}^{150}(k-3)^{2}\) using the summation formulas.

Solution:

To find: The given sum using the summation formulas.

\( \begin{align} &\sum_{k=1}^{150}(k-3)^{2} \\[0.2cm]&= \sum_{k=1}^{150} (k^2 -6k+9)\\[0.2cm] &= \sum_{k=1}^{150} k^2 - 6 \sum_{k=1}^{150} k + 9 \sum_{k=1}^{150} 1 \\[0.2cm] &= \dfrac{150(150+1)(2(150)+1)}{6}- 6 \cdot \dfrac{150(150+1)}{2} + 9 (150)\\[0.2cm] &= 1136275 -67950 + 1350\\[0.2cm] &=1069675 \end{align}\)

Answer: \( \sum_{k=1}^{150}(k-3)^{2}\) = 1,069,675.