Java Program for cube sum of first n natural numbers
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!
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