Program for cube sum of first n natural numbers

• Difficulty Level : Easy
• Last Updated : 31 Mar, 2021

Print the sum of series 13 + 23 + 33 + 43 + …….+ n3 till n-th term.
Examples :

Input : n = 5
Output : 225
13 + 23 + 33 + 43 + 53 = 225

Input : n = 7
Output : 784
13 + 23 + 33 + 43 + 53 +
63 + 73 = 784

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

A simple solution is to one by one add terms.

C++

 // Simple C++ program to find sum of series// with cubes of first n natural numbers#include using namespace std; /* Returns the sum of series */int sumOfSeries(int n){    int sum = 0;    for (int x = 1; x <= n; x++)        sum += x * x * x;    return sum;} // Driver Functionint main(){    int n = 5;    cout << sumOfSeries(n);    return 0;}

Java

 // Simple Java program to find sum of series// with cubes of first n natural numbers import java.util.*;import java.lang.*;class GFG {     /* Returns the sum of series */    public static int sumOfSeries(int n)    {        int sum = 0;        for (int x = 1; x <= n; x++)            sum += x * x * x;        return sum;    }     // Driver Function    public static void main(String[] args)    {        int n = 5;        System.out.println(sumOfSeries(n));    }} // Code Contributed by Mohit Gupta_OMG <(0_o)>

Python3

 # Simple Python program to find sum of series# with cubes of first n natural numbers # Returns the sum of seriesdef sumOfSeries(n):    sum = 0    for i in range(1, n + 1):        sum += i * i*i             return sum   # Driver Functionn = 5print(sumOfSeries(n)) # Code Contributed by Mohit Gupta_OMG <(0_o)>

C#

 // Simple C# program to find sum of series// with cubes of first n natural numbersusing System; class GFG {    /* Returns the sum of series */    static int sumOfSeries(int n)    {        int sum = 0;        for (int x = 1; x <= n; x++)            sum += x * x * x;        return sum;    }     // Driver Function    public static void Main()    {        int n = 5;        Console.Write(sumOfSeries(n));    }}// This code is contributed by// Smitha Dinesh Semwal



Javascript



Output :

225

Time Complexity : O(n)
An efficient solution is to use direct mathematical formula which is (n ( n + 1 ) / 2) ^ 2

For n = 5 sum by formula is
(5*(5 + 1 ) / 2)) ^ 2
= (5*6/2) ^ 2
= (15) ^ 2
= 225

For n = 7, sum by formula is
(7*(7 + 1 ) / 2)) ^ 2
= (7*8/2) ^ 2
= (28) ^ 2
= 784

C++

 // A formula based C++ program to find sum// of series with cubes of first n natural// numbers#include using namespace std; int sumOfSeries(int n){    int x = (n * (n + 1) / 2);    return x * x;} // Driver Functionint main(){    int n = 5;    cout << sumOfSeries(n);    return 0;}

Java

 // A formula based Java program to find sum// of series with cubes of first n natural// numbers import java.util.*;import java.lang.*;class GFG {    /* Returns the sum of series */    public static int sumOfSeries(int n)    {        int x = (n * (n + 1) / 2);         return x * x;    }     // Driver Function    public static void main(String[] args)    {        int n = 5;        System.out.println(sumOfSeries(n));    }} // Code Contributed by Mohit Gupta_OMG <(0_o)>

Python3

 # A formula based Python program to find sum# of series with cubes of first n natural# numbers # Returns the sum of seriesdef sumOfSeries(n):    x = (n * (n + 1)  / 2)    return (int)(x * x)    # Driver Functionn = 5print(sumOfSeries(n)) # Code Contributed by Mohit Gupta_OMG <(0_o)>

C#

 // A formula based C# program to// find sum of series with cubes// of first n natural numbersusing System; class GFG {         // Returns the sum of series    public static int sumOfSeries(int n)    {        int x = (n * (n + 1) / 2);         return x * x;    }     // Driver Function    public static void Main()    {        int n = 5;                 Console.Write(sumOfSeries(n));    }} // Code Contributed by nitin mittal.



Javascript



Output:

225

Time Complexity : O(1)
How does this formula work?
We can prove the formula using mathematical induction. We can easily see that the formula holds true for n = 1 and n = 2. Let this be true for n = k-1.

Let the formula be true for n = k-1.
Sum of first (k-1) natural numbers =
[((k - 1) * k)/2]2

Sum of first k natural numbers =
= Sum of (k-1) numbers + k3
= [((k - 1) * k)/2]2 + k3
= [k2(k2 - 2k + 1) + 4k3]/4
= [k4 + 2k3 + k2]/4
= k2(k2 + 2k + 1)/4
= [k*(k+1)/2]2

The above program causes overflow, even if result is not beyond integer limit. Like previous post, we can avoid overflow upto some extent by doing division first.

C++

 // Efficient CPP program to find sum of cubes// of first n natural numbers that avoids// overflow if result is going to be withing// limits.#include using namespace std; // Returns sum of first n natural// numbersint sumOfSeries(int n){    int x;    if (n % 2 == 0)        x = (n / 2) * (n + 1);    else        x = ((n + 1) / 2) * n;    return x * x;} // Driver codeint main(){    int n = 5;    cout << sumOfSeries(n);    return 0;}

Java

 // Efficient Java program to find sum of cubes// of first n natural numbers that avoids// overflow if result is going to be withing// limits.import java.util.*;import java.lang.*;class GFG {    /* Returns the sum of series */    public static int sumOfSeries(int n)    {        int x;        if (n % 2 == 0)            x = (n / 2) * (n + 1);        else            x = ((n + 1) / 2) * n;        return x * x;    }     // Driver Function    public static void main(String[] args)    {        int n = 5;        System.out.println(sumOfSeries(n));    }}// Code Contributed by Mohit Gupta_OMG <(0_o)>

Python3

 # Efficient Python program to find sum of cubes# of first n natural numbers that avoids# overflow if result is going to be withing# limits. # Returns the sum of seriesdef sumOfSeries(n):    x = 0    if n % 2 == 0 :        x = (n / 2) * (n + 1)    else:        x = ((n + 1) / 2) * n             return (int)(x * x)   # Driver Functionn = 5print(sumOfSeries(n)) # Code Contributed by Mohit Gupta_OMG <(0_o)>

C#

 // Efficient C# program to find sum of// cubes of first n natural numbers// that avoids overflow if result is// going to be withing limits.using System; class GFG {         /* Returns the sum of series */    public static int sumOfSeries(int n)    {        int x;        if (n % 2 == 0)            x = (n / 2) * (n + 1);        else            x = ((n + 1) / 2) * n;        return x * x;    }         // Driver code    static public void Main ()    {        int n = 5;        Console.WriteLine(sumOfSeries(n));    }} // This code is contributed by Ajit.



Javascript



Output:

225

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