Write a program to print the sum of the kth power of the first N natural numbers

Natural numbers are all positive integers ranging from 1 to n or infinity. Sum of first n natural number, for any number n, sum is defined as 1+2+3—-n, which is an arithmetic series whose sum is (n * (n + 1))/2. For example, If n=3, sum of the first 3 natural numbers is 1 + 2 + 3 = 6.

Formula:

Formula to calculate the Sum of First N Natural Numbers are (n * (n + 1))/2.

Example: If n=4, sum of the first 4 natural numbers is

=> (n * (n + 1))/2

=> (4 * (4 + 1))/2 => 20/2

=> 10

Write a C program to find the sum of first n natural numbers.

1. Take a number as input. 2. Find the sum of first n natural numbers using loops, function and formula.

3. Print the final result and exit.

Sum of First N Natural Numbers in C can be found out in following ways:

Method 1: (Using For Loop)

In this approach, we find the sum of the first n natural number using a for loop and in each iteration, and add ‘i’ to the sum, where i is the variable which is used to initialize the loop and sum is the variable which stores the sum of n natural numbers.

Example:

Check this: Computer Science Books | Advanced C Programming Videos

Let n = 4 Initially sum = 0, Iterate from i=1 to i=4 using a for loop, and in each iteration add i to sum.

So sum = 1+2+3+4 = 10.

Algorithm:

Step 1: Start the program. Step 2: Input the value of n. Step 3: Initialize a for loop from 1 to n. Step 4: Print sum of first n natural numbers.

Step 5: End the Program.

Here is source code of the C program to calculate the sum of first N natural numbers using for loop. The C program is successfully compiled and run on a Linux system. The program output is also shown below.

/*
* C program to find the sum of 'N' natural numbers using for loop
*/
#include <stdio.h>
int main()
{
int n;
printf("Enter number:");
scanf("%d",&n);
int sum=0;
for(int i=1;i<=n;i++)
sum += i;
printf("sum of first %d numbers is: %d",n,sum);
return 0;
}

1. Input a number n
2. Initialize a loop from 1 to n using ‘i‘ and in each iteration add the value of ‘i‘ in the sum variable.
3. Print the sum.

Time Complexity: O(n)
In the above program, a loop is executed from 1 to n, so time complexity is O(n).

Space Complexity: O(1)
Since no auxiliary space is required, space complexity is O(1).

To find the sum of the first n natural numbers, we enter the value “4” as input.

Enter number: 4 sum of first 4 numbers is: 10

Method 2: (Using While Loop)

In this approach, we find the sum of the first n natural number using a while loop. Initially i=1, execute loop until i<=n and in each iteration add i to sum and increment i.

Example:

Let n = 5 Initially, sum = 0 and i = 1

Iterate until i<=5 using a while loop and in each iteration add i to sum.

So sum = 1+2+3+4+5=15

Algorithm:

Step 1: Start the program. Step 2: Input the value of n. Step 3: Initialize a variable i and run a while loop from 1 to n. Step 4: Print sum of first n natural numbers.

Step 5: End the Program.

Here is source code of the C program to calculate the sum of first N natural numbers using while loop. The C program is successfully compiled and run on a Linux system. The program output is also shown below.

/*
* C program to find the sum of 'N' natural numbers using while loop
*/
#include <stdio.h>
int main()
{
int n;
printf("Enter number:");
scanf("%d",&n);
int sum=0,i=1;
while(i<=n)
sum += i++;
printf("sum of first %d numbers is: %d",n,sum);
return 0;
}

1. Input a number n
2. Initialize a variable i=1 and sum=0.
3. After that iterate until i is less than n using a while loop and in each iteration add i to the sum.
4. After the loop is executed, print the sum.

Time Complexity: O(n)
In the above program, a loop is executed from 1 to n, so time complexity is O(n).

Space Complexity: O(1)
Since no auxiliary space is required, space complexity is O(1).

To find the sum of the first n natural numbers, we enter the value “5” as input.

Enter number: 5 sum of first 5 numbers is: 15

Method 3: (Using Do While Loop)

In this approach, we find the sum of the first n natural number using a do while loop. Initialize a variable i and run a Do while loop from 1 to n. After the loop is executed, sum contains the sum of first n natural numbers.

Example:

Let n=6 Initially, sum = 0 and i=1,

Iterate until i<=6 using a do while loop and in each iteration add i to sum.

So sum = 1+2+3+4+5+6=21

Here is source code of the C program to calculate the sum of first N natural numbers using do while loop. The C program is successfully compiled and run on a Linux system. The program output is also shown below.

/*
* C program to find the sum of 'N' natural numbers using do while loop
*/
#include <stdio.h>
int main()
{
int n;
printf("Enter number:");
scanf("%d",&n);
int sum=0,i=1;
do
{
sum += i++;
}
while(i<=n);
printf("sum of first %d numbers is: %d",n,sum);
return 0;
}

1. Input a number n
2. Initialize a variable i=1 and sum=0.
3. After that iterate until i is less than n using a do while loop and in each iteration add i to the sum.
4. After the loop is executed, print the sum.

Time Complexity: O(n)
In the above program, a loop is executed from 1 to n, so time complexity is O(n).

Space Complexity: O(1)
Since no auxiliary space is required, space complexity is O(1).

To find the sum of the first n natural numbers, we enter the value “6” as input.

Enter number: 6 sum of first 6 numbers is: 21

Method 4: (Using Formula)

For any integer n, the sum is 1+2+3 —— n. This corresponds to an arithmetic series whose sum is (n *(n+1))/2.

So in this approach, sum is calculated using the formula: (n *(n+1))/2.

Example:

If n=7, sum = (7 * (7 + 1))/2 = 56/2 = 28
If n=3, sum = (3 * (3 + 1))/2 = 12/2 = 6

Here is source code of the C program to calculate the sum of first N natural numbers using formula. The C program is successfully compiled and run on a Linux system. The program output is also shown below.

/*
* C program to find the sum of 'N' natural numbers using formula
*/
#include <stdio.h>
int main()
{
int n;
printf("Enter number:");
scanf("%d",&n);
int sum= ((n)*(n+1))/2;
printf("sum of first %d numbers is: %d",n,sum);
return 0;
}

1. Input a number n
2. Now calculate sum of first n natural numbers by using the formula (n * (n + 1))/2 and store it in a variable sum.
3. Print the value of sum.

Time Complexity: O(1)
In the above program, only basic input output operations are performed, so time complexity is O(1).

Space Complexity: O(1)
Since no auxiliary space is required, space complexity is O(1).

To find the sum of the first n natural numbers, we enter the value “7” as input.

Enter number: 7 sum of first 7 numbers is: 28

Method 5: (Using Recursion)

The prototype of the recursive function is: int sum(int n).

Following is the function definition:

int sum(int n) { if(n==1) return 1; else return n + sum(n-1); }

Example:

Now let n=5; So function calls would be as follows.

sum(5) = 5 + sum(4) —— ——— ——— ——— ——— ——— 15 sum(4) = 4 + sum(3) ——— ——— ————— —— 10 sum (3) = 3 + sum(2) ——— ——— —- 6 sum(2) = 2 + sum(1) ——— 3 sum(1) = 1 // Base Case

Algorithm:

Step 1: Start the program. Step 2: Input the value of n.

Step 3: Declare a function func to find sum.

Base case: When n=1,return 1
Other cases: return n + func(n-1)

Step 4: Print the value of sum as returned by the func.
Step 5: End the Program

Here is source code of the C program to calculate the sum of first N natural numbers using recursion. The C program is successfully compiled and run on a Linux system. The program output is also shown below.

/*
* C program to find the sum of 'N' natural numbers using Recursion.
*/
#include <stdio.h>
int sum(int n)
{
//Base Case
if(n==1)
return 1;
return n + sum(n-1);
}
int main()
{
int n;
printf("Enter number: ");
scanf("%d",&n);
printf("Sum of first %d natural number is: %d",n,sum(n));
return 0;
}

1. Input a number n
2. Declare a function sum which takes ‘n‘ as the parameter with base condition to stop when n is equal to 1.
3. Print the value returned by function as the sum.

Time Complexity: O(n)
The program above has n function calls, so the time complexity is O(n).

Space Complexity: O(1)
Since no auxiliary space is required, space complexity is O(1).

To find the sum of the first n natural numbers, we enter the value “5” as input.

Enter number: 5 sum of first 5 numbers is: 15

Method 6: (Using Function)

The prototype of the recursive function is: int sum_n(int n). It takes n as the parameter and returns the sum of first n natural numbers.

Algorithm:

Step 1: Start the program. Step 2: Input the value of n. Step 3: Declare a function func to find sum. Step 4: Print the value of sum as returned by the func.

Step 5: End the Program.

Here is source code of the C program to calculate the sum of first N natural numbers using function. The C program is successfully compiled and run on a Linux system. The program output is also shown below.

/*
* C program to find the sum of 'N' natural numbers using function.
*/
#include <stdio.h>
int sum_n(int n)
{
int s=0;
for(int i=1;i<=n;i++)
s+=i;
return s;
}
int main()
{
int n;
printf("Enter n:");
scanf("%d",&n);
printf("sum of first %d numbers is: %d",n,sum_n(n));
return 0;
}

1. Input a number n
2. Declare a function sum which takes ‘n‘ and return the sum of first n natural numbers.
3. Print the value returned by function sum.

Time Complexity: O(n)
The program above has n function calls, so the time complexity is O(n).

Space Complexity: O(1)
Since no auxiliary space is required, space complexity is O(1).

To find the sum of the first n natural numbers, we enter the value “120” as input.

Enter number: 15 sum of first 15 numbers is: 120

In this approach, we store the natural numbers in an array then find the sum of the array which is equal the sum of first n natural numbers.

Example:

If n=5;
So array of first 5 natural number is: [1,2,3,4,5]

Sum of array = 1+2+3+4+5=15;

Sum of first 5 natural numbers=sum of array=15

Here is source code of the C program to calculate the sum of first N natural numbers using array. The C program is successfully compiled and run on a Linux system. The program output is also shown below.

/*
* C program to find the sum of 'N' natural numbers using array.
*/
#include <stdio.h>
int main()
{
int n;
printf("Enter number:");
scanf("%d",&n);
int arr[n];
// Store n natural numbers in an array
for(int i=0;i<n;i++)
arr[i] = i +1 ;
int sum=0;
for(int i=0;i<n;i++)
sum+=arr[i];
printf("sum of first %d numbers is: %d",n,sum);
return 0;
}

1. Input a number n
2. Store first n natural numbers in an array.
3. Find the sum of the array and store it in sum.
4. Print the value of sum.

Time Complexity: O(n)
In the above program, loop is executed from 1 to n, so time complexity is O(n).

Space Complexity: O(1)
Since no auxiliary space is required, space complexity is O(1).

To find the sum of the first n natural numbers, we enter the value “5” as input.

Enter number: 5 sum of first 5 numbers is: 15

Method 8: Sum of N Natural Numbers in a Given Range

In this approach we traverse between the intervals using the iterative statement to find the sum of natural numbers in the interval.

Example:

Let low = 2; high = 9;

So sum = 2+3+4+5+6+7+8+9 (from low to high)

sum = 44.

Here is source code of the C program to calculate the sum of N natural numbers in a Given Range. The C program is successfully compiled and run on a Linux system. The program output is also shown below.

/*
* C program to find the sum of 'N' natural numbers in a Given Range.
*/
#include <stdio.h>
int main()
{
int l,r;
printf("Enter ends of interval to find sum:");
scanf("%d %d",&l,&r);
int sum=0;
for(int i=l;i<=r;i++)
sum +=i;
printf("Sum is: %d",sum);
return 0;
}

1. Input the two ends of the interval i.e high and low.
2. Initialize sum = 0 and initiate a loop from low to high using i.
3. Add i to sum in each iteration.
4. Then Print the sum variable.

Time Complexity: O(n)
In the above program, since there is a loop executed from low to high, the time complexity is linear.

Space Complexity: O(1)
Since no auxiliary space is required, space complexity is O(1).

To find the sum of n natural numbers in a given range, we enter the values low = “2” and high = “9” as input.

Enter ends of interval to find sum:2 9 Sum is: 44

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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