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
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 <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 Function int 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 series def sumOfSeries(n): sum = 0 for i in range ( 1 , n + 1 ): sum + = i * i * i return sum # Driver Function n = 5 print (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 numbers using 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
<?php // Simple PHP program to find sum of series // with cubes of first n natural numbers // Returns the sum of series function 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. ?> |
Javascript
<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++
// 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 Function int 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 series def sumOfSeries(n): x = (n * (n + 1 ) / 2 ) return ( int )(x * x) # Driver Function n = 5 print (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 numbers using 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
<?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. ?> |
Javascript
<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++
// 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 // numbers 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 int 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 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)> |
Python3
# 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 series def sumOfSeries(n): x = 0 if n % 2 = = 0 : x = (n / 2 ) * (n + 1 ) else : x = ((n + 1 ) / 2 ) * n return ( int )(x * x) # Driver Function n = 5 print (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 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
<?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 numbers function 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. ?> |
Javascript
<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)
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