# 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?

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?

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

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