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Updated on 14th February, 2023 , 4 min read
Using the formula for the sum of all natural numbers as well as arithmetic progression, it is simple to calculate the sum of even numbers from 2 to infinity. We are aware that even numbers are those that can be divided evenly by two. They are 2, 4, 6, 8, 10, 12, 14, 16, and so on. We now need to calculate the sum of these numbers.
The numbers beginning at 2 and continuing until infinity are the sum of even numbers. As we already know, even numbers are those that can be divided by 2, such as 2, 4, 6, 8, 10, and so forth. The sum of the arithmetic progression formula or the sum of natural numbers formula must be used to calculate the sum of even numbers.
The formula for calculating the arithmetic progression is used to determine the formula for the sum of even numbers. Even numbers can be added up indefinitely. The formula for the sum of natural numbers can also be used to evaluate the sum of even numbers. We need to obtain the formula for 2 + 4 + 6 + 8 + 10 +...... 2n. The sum of even numbers = 2(1 + 2+ 3+ .....n). This implies 2(sum of n natural numbers) = 2[n(n+1)]/2 = n(n+1)
S = n(n+1) , where n is the number of terms in the series
Let's use arithmetic progression to arrive at the formula for the sum of even numbers. Let the sum of the first n even numbers be Sn, so Sn = 2 + 4+6+8+10+ …………………..+(2n) ……. (1)
By arithmetic progression (AP), we know that, for any sequence, the sum of n terms of an AP is given by: Sn = (1/2)× n[2a+(n-1)d] ……..(2)
Where,
n = number of terms in the series
a = First term of an arithmetic progression
d= Common difference in an arithmetic progression
Therefore, if we put the values in equation 2 with respect to equation 1, such as a = 2, d = 2, and suppose last term, l = (2n)
So, the sum will be:
Sn = ½ n[2×2+(n-1)2]
Sn = ½ n[4+2n-2]
Sn = ½ n[2+2n]
Sn = ½ 2n(n+1)
Sn = n(n+1)
Hence, the sum of even numbers formula = n(n+1)
The sum of 1 to 10 consecutive even numbers is shown in the table below.
|
Number of consecutive even numbers (n) |
Sum of even numbers (Sn = n (n+1)) |
Recheck |
|
1 |
1(1+1)=1×2=2 |
2 |
|
2 |
2(2+1) = 2×3 = 6 |
2+4 = 6 |
|
3 |
3(3+1)=3×4 = 12 |
2+4+6 = 12 |
|
4 |
4(4+1) = 4 x 5 = 20 |
2+4+6+8=20 |
|
5 |
5(5+1) = 5 x 6 = 30 |
2+4+6+8+10 = 30 |
|
6 |
6(6+1) = 6 x 7 = 42 |
2+4+6+8+10+12 = 42 |
|
7 |
7(7+1) = 7×8 = 56 |
2+4+6+8+10+12+14 = 56 |
|
8 |
8(8+1) = 8 x 9 = 72 |
2+4+6+8+10+12+14+16=72 |
|
9 |
9(9+1) = 9 x 10 = 90 |
2+4+6+8+10+12+14+16+18=90 |
|
10 |
10(10+1) = 10 x 11 =110 |
2+4+6+8+10+12+14+16+18+20=110 |
Find the first ten even numbers by counting them. The following even numbers will be on the list of the first even numbers: 2, 4, 6, 8, 10, 12, 14, 16, and 18.
The sum of consecutive even numbers from 1 to 10 is therefore Sn = 2 + 4 + 6 + 8 + 10 +... 10 terms.
S = 10(10+1) = 10 x 11 = 110, where n = 10, according to the formula Sn = n(n+1).
Also, 2+4+6+8+10+12+14+16+18+20=110
thus checked.
We are aware that even numbers are those that can be divided by two. Additionally, we are aware that there is a 2 difference between any two consecutive even numbers. The total of all the even numbers in the list from 1 to 100 can be calculated by adding up the even numbers from 1 to 100. There are 50 even numbers between 1 and 100 according to the definition of even numbers. Thus, n = 50
In the equation for the sum of even numbers, Sn = n(n+1), replace the value of n.
Consequently, Sn = 50(50+1) = 50 x 51 = 2550.
The total of all the even numbers in the list from 1 to 50 can be calculated by adding up the even numbers from 1 to 50. Even numbers are those that fall into the categories of 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48, and 50. Thus, n = 25.
The values in the formula Sn = n(n+1) should be substituted.
Therefore, S = 25(25+1) = 25 x 26 = 650
The total of all the even numbers in the list from 51 to 100 can be calculated by adding up the even numbers from 51 to 100. The even numbers from 51 to 100 are 52, 54, 56, 58, 60, 62, 64, 66, 68, 70, 72, 74, 76, 78, 80, 82, 84, 86, 88, 90, 92, 94, 96, 98, and 100. There are thus 25 even numbers between 51 and 100.
Here, n = 25, d = 2, and a = 52.
Using the sum-of-ap formula,
Sn=1/2×n[2a+(n-1)d]
S=1/2×25[2.52+(25-1)2]
S=1/2×25[104+(24)2]
S=25/2[152]
S=[25(76)] = 1900
The following are some crucial details:
Example: Determine the sum of even numbers from 1 to 100.
Solution: We are aware that there are 50 even numbers, from 1 to 100,.
So, n = 50
Using the sum of even numbers formula i.e..
Sn= n (n+1)
Sn= 50 (50+1)
= 50 x 51 = 2550
Example: What is the sum of even numbers from 1 to 200?
Solution:We are familiar that, from 1 to 200, there are in total 100 even numbers.
Therefore, n =100
Using the formula for sum of even numbers we know;
Sn= n (n+1)
Sn= 100 (100+1)
= 100 x 101
= 10100
Example: Evaluate the sum of squares of odd numbers.
Solution:Sum of squares of 3 odd numbers = n (2n+1) (2n-1) ÷5
= 5(2 x 5+1) (2 x 5-1) ÷ 5
= 5 (11) (9) ÷ 5
= 99
Example: Determine the sum of the squares of the first three odd numbers.
Solution: Sum of squares of three odd numbers = n (2n+1) (2n-1) ÷3
= 3 (2 x 3 + 1) (2 x 3-1) ÷ 3
= 5 (7) (5) / 5
= 35
To Prove = 1² + 5² + 7²
= 1 + 25 + 49
= 35
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Ans. Between 1 and 100, there are only even numbers: 2, 4, 6, 8, 98, and 100. l = 100, a = 2, n = ? Between 1 and 100, the total of all even numbers is 2550.
Ans. The sum of the first 12 even numbers is 156.
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0 is an even number.
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