Last Updated: 25 Nov, 2020
Sum Of Even Numbers Till N
Easy
Asked in companies
You have been given a number 'N'. Your task is to find the sum of all even numbers from 1 to 'N' (both inclusive).
Example :
Given 'N' : 6
Sum of all even numbers till 'N' will be : 2 + 4 + 6 = 12
Input Format :
The first line of the input contains an integer 'T', denoting the number of test cases.
The first line and the only line of each test case contains an integer 'N'.
Output Format :
For each test case, print a single integer representing the sum of even numbers till 'N'.
Print a single line for each test case.
Note:
You do not need to print anything, it has already been taken care of. Just implement the given function.
Constraints :
1 <= T <= 10
1 <= N <= 10^5
Time Limit: 1 sec
Approaches
We can simply apply brute force and add numbers to a variable.
Here is the algorithm :
- Create a variable (say, ‘SUM’) which will store the sum of even numbers and initialise it to 0.
- Create a variable (say, ‘i’ ) which we will use to iterate and initialise it to 2 (first even number).
- Run a loop till i is smaller then equal to ‘N’ and do :
- Update ‘SUM’ to ‘SUM’ + ‘i’
- Update ‘i’ to ‘i’ + 2 (We increment 2 times so that we can reach to next even number)
- Finally, return ‘SUM’
We can derive a formula to calculate the sum of even numbers till ‘N’.
Let the number given be : ‘N’. Then the even numbers need to be added will be {2, 4, 6,…., ‘N or ‘N - 1’}. Total even numbers from 1 to ’N' will be ‘N/2’. Example :
‘N’ : 4, even numbers till 4 : {2, 4}, count = ‘N/2’ = 2
‘N’ : 5, even numbers till 5 : {2, 4}, count = ‘N/2’ = 2
‘N’ : 1, even numbers till 1 : {}, count = ‘N/2’ = 0
The sequence of even numbers till ‘N’ forms an A.P. (arithmetic progression) with common difference, ‘D’ as 2, first element, ‘A’ as 2 and number of elements, ‘N’ as ‘N/2’ (as proved above).
Sum of A.P. is : (N/2) * (A + L) where ‘N’ is number of terms, ‘A’ is first number and ‘L’ is last number which can also be written as ‘A + (N - 1)D'.
Putting ‘N’ = ‘N/2’, ‘A’ = 2 and ‘D’ = 2 in above formula,
‘SUM' = (N/2*2) * (2 + 2 + (N/2 - 1)*2)
= (N/2) * (1 + 1 + N/2 - 1)
= (N/2) * (N/2 + 1)
We can calculate the sum using above formula directly.