Bash Script to Find Sum of Odd Numbers

Use a Bash loop with an if condition to check odd numbers and add them to a sum variable, like: sum=0; for i in {1..N}; do if (( i % 2 != 0 )); then sum=$((sum + i)); fi; done; echo $sum.

Examples

Input5

Output9

Input10

Output25

Input1

Output1

How to Think About It

To find the sum of odd numbers up to a limit, think of counting from 1 to that number. For each number, check if it is odd by using the remainder operator %. If the remainder when divided by 2 is not zero, add it to a running total. Finally, print the total sum.

Algorithm

1

Get the input number N from the user

2

Initialize a sum variable to 0

3

Loop from 1 to N

4

Check if the current number is odd using the remainder operator

5

If odd, add it to sum

6

After the loop ends, print the sum

Code

bash

read -p "Enter a number: " N
sum=0
for (( i=1; i<=N; i++ ))
do
  if (( i % 2 != 0 ))
  then
    sum=$((sum + i))
  fi
done
echo "$sum"

Output

9

Dry Run

Let's trace input 5 through the code

1

Initialize sum

sum=0

2

Loop i=1

1 % 2 != 0 is true, sum=0+1=1

3

Loop i=2

2 % 2 != 0 is false, sum=1

4

Loop i=3

3 % 2 != 0 is true, sum=1+3=4

5

Loop i=4

4 % 2 != 0 is false, sum=4

6

Loop i=5

5 % 2 != 0 is true, sum=4+5=9

i	Is Odd?	Sum
1	Yes	1
2	No	1
3	Yes	4
4	No	4
5	Yes	9

Why This Works

Step 1: Check odd numbers

The code uses i % 2 != 0 to find odd numbers because odd numbers leave a remainder of 1 when divided by 2.

Step 2: Add to sum

When a number is odd, it is added to the sum variable to keep a running total.

Step 3: Print result

After checking all numbers, the script prints the total sum of all odd numbers found.

Alternative Approaches

Using while loop

bash

read -p "Enter a number: " N
sum=0
i=1
while (( i <= N ))
do
  if (( i % 2 != 0 ))
  then
    sum=$((sum + i))
  fi
  ((i++))
done
echo "$sum"

This uses a while loop instead of for loop; functionally the same but some prefer while for clarity.

Using arithmetic progression formula

bash

read -p "Enter a number: " N
max_odd=$(( N % 2 == 0 ? N - 1 : N ))
count=$(((max_odd + 1) / 2))
sum=$(( count * count ))
echo "$sum"

This uses the formula for sum of first k odd numbers (k^2) for faster calculation without looping.

Complexity: O(n) time, O(1) space

Time Complexity

The script loops through all numbers from 1 to N once, so it runs in linear time O(n).

Space Complexity

Only a few variables are used regardless of input size, so space complexity is O(1).

Which Approach is Fastest?

The arithmetic formula approach is fastest with O(1) time, but looping is simpler and more intuitive for beginners.

Approach	Time	Space	Best For
For loop with condition	O(n)	O(1)	Simple and clear for beginners
While loop with condition	O(n)	O(1)	Alternative loop style
Arithmetic formula	O(1)	O(1)	Fastest for large inputs, less intuitive