How to write a C Program to Find Sum of series 1²+2²+3²+….+n² using For Loop, and Functions with example. Before we get into the example, The Mathematical formula for C Program to find Sum of series 1²+2²+3²+….+n² = ( n (n+1) (2n+1)) / 6
C Program to Find Sum of series 1²+2²+3²+….+n²
In this C program, the user asked to enter any positive integer. Then using that value, the compiler will find the sum of series 12 + 22 + 32 + … + n2 using the above formula.
Within the main() function, We declared 2 integer variables Number and Sum. The printf statement will ask the user to enter any integer value. And the below C Programming scanf statement will assign the user integer value to the variable name Number
/* C Program to Calculate Sum of series 1²+2²+3²+....+n² */ #include <stdio.h> int main() { int Number, Sum = 0; printf("\n Please Enter any positive integer \n"); scanf(" %d",&Number); Sum = (Number * (Number + 1) * (2 * Number + 1 )) / 6; printf("\n The Sum of Series for %d = %d ",Number, Sum); }
Please Enter any positive integer
5
The Sum of Series for 5 = 55
In the next line, We are calculating the Sum of the series 1²+2²+3²+4²+5² using above formula
Sum = (Number * (Number + 1) * (2 * Number + 1 )) / 6;
Sum = (5 * (5 + 1) * (2 * 5 +1)) / 6
Sum = (5 * 6 * 11) / 6
Sum = 330 /6
Sum = 55
The C Program to Find Sum of series 1²+2²+3²+….+n² final output for 5 = 55
C Program to calculate Sum of series 1²+2²+3²+….+n²
If you want to display the series order 12 + 22 + 32 +42 + 52 in the program output, then we have to add extra For loop to display as below
#include <stdio.h> int main() { int Number, i, Sum = 0; printf("\nPlease Enter any positive integer \n"); scanf("%d",&Number); Sum = (Number * (Number + 1) * (2 * Number + 1 )) / 6; for(i =1; i<=Number;i++) { if (i != Number) printf("%d^2 + ",i); else printf("%d^2 = %d ",i, Sum); } }
The For loop inside the main function will traverse each member and displays the output.
In the above screenshot, the user entered value is 4 So,
Sum = (Number * (Number + 1) * (2 * Number + 1 )) / 6;
Sum = (4 * (4 + 1) * (2 * 4 +1))/6
Sum = (4 * 5 * 9) / 6
= 180 /6
Sum = 30
Now, the compiler will enter into for loop
First Iteration
i = 1, so the condition inside the for loop (i <= Number) is TRUE (1 <=4). Next, It will go to if condition (i != Number). It means (1 != 4) – Which is TRUE. So, it will print the output as 1²+
i = 2. Repeat the same until i reaches 4. When i = 4, if condition fails. So, the Else statement is printed.
The final Output will be 1²+2²+3²+4² = 30
C Program to Find Sum of series 1²+2²+3²+….+n² using Functions
In this C program, the user enters any positive integer and then using that value, compiler will find the sum of series 12 + 22 + 32 + … + n2 using functions.
#include <stdio.h> void Sum_Of_Series(int); int main() { int Number; printf("\n Please Enter any positive integer \n"); scanf("%d",&Number); Sum_Of_Series(Number); } void Sum_Of_Series(int Number) { int i, Sum; Sum = (Number * (Number + 1) * (2 * Number + 1 )) / 6; for(i =1;i<=Number;i++) { if (i != Number) printf("%d^2 + ",i); else printf(" %d^2 = %d ", i, Sum); } }
Please Enter any positive integer
9
1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2 + 7^2 + 8^2 + 9^2 = 285
Program to Find Sum of series 1²+2²+3²+….+n² using Recursion
Please refer to Find Sum of series 1²+2²+3²+….+n² using Recursion in Recursion Article to see the code and working principle.
Comments are closed.