In this article we will show you, How to write a C Program to Find Sum of series 1³+2³+3³+….+n³ using For Loop, Recursion, and Functions with example. This series is also called as Sum of cubes of n natural numbers. The Mathematical formula for Sum of series 1³+2³+3³+….+n³ = ( n (n+1) / 6)²
C Program to Find Sum of series 1³+2³+3³+….+n³
In this C program user is asked to enter any positive integer and then using that value, compiler will find the sum of series 1³+2³+3³+….+n³ using the above formula.
CODE
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/* Sum of Cubes of n Natural Numbers */ #include <stdio.h> #include <math.h> int main() { int Number, Sum = 0; printf("\n Please Enter any positive integer \n"); scanf(" %d",&Number); Sum = pow(((Number * (Number + 1))/ 2), 2); printf("\n The Sum of Series for %d = %d ",Number, Sum); } |
OUTPUT
ANALYSIS
Within the main() function,
We declared 2 integer variables Number and Sum. Below printf statement will ask the user to enter any integer value.
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printf("\n Please Enter any positive integer \n"); |
Below scaf statement wil assign the user integer value to the variable name Number
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scanf(" %d",&Number); |
Sum = pow (((Number * (Number + 1)) / 2), 2);
Sum = pow (((5 * (5 + 1)) / 2), 2);
Sum = pow (((5 * 6) / 2), 2);
Sum = pow ((30 / 2), 2);
Sum = pow (15, 2);
Sum = 15 * 15 = 225
Final printf statement will print the Sum as output
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printf("\n The Sum of Series for %d = %d ",Number, Sum); |
The Sum of Series for 5 = 225
C Program to Find Sum of series 1³+2³+3³+….+n³ using For Loop
If you want to display the series order 1³+2³+3³+4³ in the output then we have to add extra For Loop in C Programmingto display as below
CODE
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/* Sum of Cubes of n Natural Numbers */ #include <stdio.h> #include <math.h> int main() { int Number, i, Sum = 0; printf("\nPlease Enter any positive integer \n"); scanf("%d",&Number); Sum = pow(((Number * (Number + 1))/ 2), 2); for(i =1; i<=Number;i++) { if (i != Number) printf("%d^3 + ",i); else printf("%d^3 = %d ",i, Sum); } } |
OUTPUT
ANALYSIS
We declared 3 integer variables Number, i and Sum. For loop inside the main function will traverse each and every member and displays the output
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for(i =1; i<=Number;i++) { if (i != Number) printf(" %d^3 + ",i); else printf(" %d^3 = %d ",i, Sum); } |
Sum = 100
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 1³+
i is incremented to 2.
It will do the same until i reaches 4. When it reached 4 then if condition will fails so, Else statement is printed.
The final Output will be 1³+2³+3³+4³ = 100
C Program to Find Sum of series 1³+2³+3³+….+n³ using Functions
In this C program user is asked to enter any positive integer and then using that value, compiler will find the sum of series 1³+2³+3³+….+n³ using FUNCTIONS.
CODE
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/* Sum of Cubes of n Natural Numbers */ #include <stdio.h> #include <math.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 = 0; Sum = pow (((Number * (Number + 1))/ 2), 2);; for(i =1;i<=Number;i++) { if (i != Number) printf("%d^3 + ",i); else printf(" %d^3 = %d ", i, Sum); } } |
OUTPUT
ANALYSIS
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void Sum_Of_Series(int); |
Within the main() function,
We declared 1 integer variable Number. Below printf statement will ask the user to enter any integer value.
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printf("\n Please Enter any positive integer \n"); |
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scanf("%d",&Number); |
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Sum_Of_Series(Number); |
When the compiler reaches the function calling then it will jump to the function definition for the calculations. We already explained the logic in the above example.
C Program to Find Sum of series 1³+2³+3³+….+n³ using Recursion
This C program allows the user to enter the value of N then, it will find the s sum of series 1³+2³+3³+….+n³ using RECURSION
CODE
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/* Sum of Cubes of n Natural Numbers */ #include <stdio.h> int Sum_Of_Series(int); int main() { int Number, Sum; printf("\nPlease Enter any positive integer \n"); scanf("%d",&Number); Sum =Sum_Of_Series(Number); printf("\nSum of the Series = %d",Sum); } int Sum_Of_Series(int Number) { if(Number == 0) return 0; else //Recursive Calling return (Number * Number * Number) + Sum_Of_Series(Number-1); } |
OUTPUT
ANALYSIS
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int Sum_Of_Series(int); |
First line of the program is the declaration of User Defined Function
Within the main() function,
We declared 2 integer variables Number and Sum. In the next line, we called the user defined function Sum_Of_Series() and assigned it to the integer variable Sum.
Sum = Sum_Of_Series(Number);
When the compiler reaches the function calling then it will jump to the function definition for the calculations.
Function Definition
Within the Sum_Of_Series (Number) function,
If the user entered Number is 0 then the function will return 0 else it will return
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(Number * Number * Number) + Sum_Of_Series(Number-1); |
Let us divide the above expression for better understanding
(Number * Number * Number) = Multiplying the number three times
Sum_Of_Series(Number-1) = Calling the same function with 1 number minus
From the above output,
User entered value is 3,
Recursion 1
Number = 3 which is Greater than 0 and Sum is 0 so,
Sum = (Number * Number) + Sum_Of_Series(Number-1)
Sum = (3 * 3 * 3) + Sum_Of_Series (3 – 1)
Sum = 27 + Sum_Of_Series (2)
Sum value is = 27
Recursion 2
Number = 2 which is Greater than 0 and Sum is 27 so,
Sum = (Number * Number) + Sum_Of_Series (Number-1)
Sum = (2 * 2 * 2) + Sum_Of_Series (2 – 1)
Sum = 8 + Sum_Of_Series (1)
Sum value is: 27 + 8 = 35
Recursion 3
Number = 1 which is Greater than 0 and Sum is 35 so,
Sum = (Number * Number) + Sum_Of_Series (Number-1)
Sum = (1 * 1 * 1) + Sum_Of_Series (1 – 1)
Sum = 1 + Sum_Of_Series (0)
Sum value is: 35 + 1 = 36
Recursion 4
Number = 0 which means First if condition is True so it will exit from the function. Final value of Sum is 36.
NOTE: We must use some sort condition to exit the recursive calling. If you forgot the condition then the function will execute infinite times.
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