Java Program for cube sum of first n natural numbers

17 Mar, 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
Java
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)>
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
Java
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)>
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. 
Java
Java
// Efficient Java program to find sum of cubes// of first n natural numbers that avoids// overflow if result is going to be within// 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)>
Output: 
225
Please refer complete article on Program for cube sum of first n natural numbers for more details! 
surinderdawra388
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