If a *whole number* is divisible by 2 into whole numbers then those numbers are called **even numbers**. Example: **2, 4, 6, 8,** and so on… are even numbers. But, what if you want to find the sum of first n even numbers? What formula will you use to do so?

This article is going to explain to you the method and formula to find that sum and also the **sum of first n even numbers**.

**Related Topic:** Sum of Odd Numbers Formula

Index

**General Formula for Even Numbers**

As we know, the Even numbers are

2, 4, 6, 8, 10, …

If we observe carefully, we can find an Arithmetic Progression Sequence(AP),

Here, the AP can be made into the formula as,

The formula for Even numbers = \(\)2n\(\)

## Sum of Even Numbers

Let us say the sequence of the n even numbers is

=> 2k + 2(k+1) + …… + 2(k+n-1)

From Arithmetic Progression, we know that,

\(S_n = \frac{n}{2}[2a + (n-1)d] \mbox{ OR } \frac{n}{2}[a + l]\)

Where,

\(n\) = no. of terms

\(a\) = First term of the A.P.

\(d\) = Common difference

\(l\) = Last term

And we know,

Here, a = 2k, d = 2, n = n & l = 2(k+n-1)

So, the sum would be \(S_n = \frac{n}{2}[2(2k) + (n-1)2]\)

= \(S_n = \frac{n}{2}[4k + 2n -2 ]\) = \(S_n = (n)[2k -1 + n]\)

**Sum of **F**irst n Even Numbers**

Lets find the sum of first \(n\) Even numbers: (\(S_n\)).

Therefore, \(S_n = 2 + 4 + 6 + 8 ….+ 2n\)

\(S_n = \sum_{i = 1} ^{n} (2i)\)

From Arithmetic Progression, we know that,

\(S_n = \frac{n}{2}[2a + (n-1)d] \mbox{ OR } \frac{n}{2}[a + l]\)

Where,

\(n\) = no. of terms

\(a\) = First term of the A.P.

\(d\) = Common difference

\(l\) = Last term

Here, \(a = 2, d = 2 \mbox{ & } l = 2n\)

Therefore, \(S_n = \frac{n}{2} [a + l]\)

\(S_n = \frac{n}{2} [2 + 2n] = n [1+n]\)

Therefore, \(S_n = n(n+1)\)

For,

- n = 1,
- \(S_1 = 2\) (&)
- \(S_1 = 1 [ 1 + 1] = 2\)

- n = 2,
- \(S_2 = 2 + 4 = 6\) (&)
- \(S_2 = 2[ 1 + 2] = 6\)

- n = 3,
- \(S_3 = 2 + 4 + 6 = 12\) (&)
- \(S_3 = 3[ 1 + 3] = 12\)

- n = 4,
- \(S_4 = 2 + 4 + 6 + 8 = 20\) (&)
- \(S_4 = 4[ 1 + 4] = 20\)

and so on… .

**Solved Examples**

**Question 1.** Find sum of first 50 Even numbers.

**Solution.** \(S_n = n[1 + n]\)

Therefore, \(S_{50} = 50[1 + 50] = 2550\)

**Question 2.** Find the sum of Even number between 50 & 100.

**Solution.** \(S_n = n[1 + n]\)

Therefore, sum of Even number between 50 & 100 = \(S_{50} – S_{25}\)

[Because there are 25 Even number in 1 to 50 & 50 Even numbers in between 1 to 100]

\(S_{25} = 25[1 + 25] = 650\)

\(S_{50} = 50[1 + 50] = 2550\)

Therefore, sum of Even number between 50 & 100 = 2550 – 650 = 1900.

**FAQs**

**What is the sum of first even Numbers?**Sum of first ‘n’ even numbers: \(S_n = n[n + 1]\)

**What are first ten even numbers?**First 10 even numbers are 2, 4, 6, 8, 10, 12, 14, 16, 18, 20

**What are even numbers?**A whole number that is divisible by 2 into whole numbers then those numbers are Even numbers.

**What is sum of first 20 natural numbers?**\(S_{20} = \frac{n(n + 1)}{2} = \frac{20(20 + 1)}{2} = 210\).