# Sum of Even Numbers Formula

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 First 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?

$$S_{20} = \frac{n(n + 1)}{2} = \frac{20(20 + 1)}{2} = 210$$.

Scroll to Top