Program for cube sum of first n natural numbers

22 Jun, 2022
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
 
A simple solution is to one by one add terms.  
C++
Java
Python3
C#
PHP
Javascript
// Simple C++ program to find sum of series// with cubes of first n natural numbers#include <iostream>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;}
// 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)>
# 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)>
// 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
<?php// Simple PHP program to find sum of series// with cubes of first n natural numbers // Returns the sum of seriesfunction sumOfSeries( $n){    $sum = 0;    for ($x = 1; $x <= $n; $x++)        $sum += $x * $x * $x;    return $sum;} // Driver code$n = 5;echo sumOfSeries($n); // This Code is contributed by vt_m.?>
<script> //  Simple javascript program to find sum of series// with cubes of first n natural numbers /* Returns the sum of series */function sumOfSeries( n){    let sum = 0;    for (let x = 1; x <= n; x++)        sum += x * x * x;    return sum;} // Driven Program     let n = 5;     document.write(sumOfSeries(n)); // This code contributed by aashish1995 </script>
Output : 
225
Time Complexity: O(n)
Auxiliary Space: O(1)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++
Java
Python3
C#
PHP
Javascript
// A formula based C++ program to find sum// of series with cubes of first n natural// numbers#include <iostream>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;}
// 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)>
# 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)>
// 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.
<?php// A formula based PHP program to find sum// of series with cubes of first n natural// numbers function sumOfSeries($n){    $x = ($n * ($n + 1) / 2);    return $x * $x;} // Driver Function$n = 5;echo sumOfSeries($n); // This code is contributed by vt_m.?>
<script> // Simple javascript program to find sum of series// with cubes of first n natural numbers /* Returns the sum of series */function sumOfSeries( n){    x = (n * (n + 1) / 2)    return (x * x)} // Driven Program     let n = 5;    document.write(sumOfSeries(n)); // This code is contributed by sravan kumar </script>
Output:  
225
Time Complexity: O(1)
Auxiliary Space: 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++
Java
Python3
C#
PHP
Javascript
// Efficient CPP program to find sum of cubes// of first n natural numbers that avoids// overflow if result is going to be with in// limits.#include <iostream>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;}
// Efficient Java program to find sum of cubes// of first n natural numbers that avoids// overflow if result is going to be with in// 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)>
# Efficient Python program to find sum of cubes# of first n natural numbers that avoids# overflow if result is going to be with in# 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)>
// Efficient C# program to find sum of// cubes of first n natural numbers// that avoids overflow if result is// going to be with in 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.
<?php// Efficient PHP program to// find sum of cubes of first // n natural numbers that avoids// overflow if result is going// to be with in limits. // Returns sum of first n// natural numbersfunction sumOfSeries($n){    $x;    if ($n % 2 == 0)        $x = ($n / 2) * ($n + 1);    else        $x = (($n + 1) / 2) * $n;    return $x * $x;} // Driver code$n = 5;echo sumOfSeries($n); // This code is contributed by vt_m.?>
<script> // Simple javascript program to find sum of series// with cubes of first n natural numbers /* Returns the sum of series */function sumOfSeries( n){  x=0    if (n % 2 == 0)        x = (n / 2) * (n + 1)    else        x = ((n + 1) / 2) * n             return (x * x)} // Driven Program     let n = 5;    document.write(sumOfSeries(n)); // This code contributed by sravan </script>
Output:  
225
Time complexity: O(1) since performing constant operations
Auxiliary Space: O(1)This article is contributed by R_Raj. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above. 
Smitha Dinesh Semwal
nitin mittal
jit_t
vt_m
aashish1995
sravankumar8128
surinderdawra388
kumargaurav97520
maths-cube
number-theory
Mathematical
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Mathematical
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