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.
// 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;
} |
// 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 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)> |
// 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 // 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. ?> |
<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
// 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;
} |
// 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 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)> |
// 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 // 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.
// 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;
} |
// 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 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)> |
// 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 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. ?> |
<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)