Given a number x and n, the task is to find the sum of the below series of x till n terms:
Examples:
Input: x = 5, n = 2
Output: 7.67
Explanation:
Sum of first two termed
Input: x = 5, n = 4Output: 18.08Explanation:
Approach: Iterate the loop till the nth term, compute the formula in each iteration i.e.
nth term of the series =
Below is the implementation of the above approach:
// C++ Program to compute sum of // 1 + x/2! + x^2/3! +...+x^n/(n+1)! #include <iostream> #include <math.h> using namespace std;
// Method to find the factorial of a number int fact( int n)
{ if (n == 1)
return 1;
return n * fact(n - 1);
} // Method to compute the sum double sum( int x, int n)
{ double i, total = 1.0;
// Iterate the loop till n
// and compute the formula
for (i = 1; i <= n; i++) {
total = total + ( pow (x, i) / fact(i + 1));
}
return total;
} // Driver code int main()
{ // Get x and n
int x = 5, n = 4;
// Print output
cout << "Sum is: " << sum(x, n);
return 0;
} |
// Java Program to compute sum of // 1 + x/2! + x^2/3! +...+x^n/(n+1)! public class SumOfSeries {
// Method to find factorial of a number
static int fact( int n)
{
if (n == 1 )
return 1 ;
return n * fact(n - 1 );
}
// Method to compute the sum
static double sum( int x, int n)
{
double total = 1.0 ;
// Iterate the loop till n
// and compute the formula
for ( int i = 1 ; i <= n; i++) {
total = total + (Math.pow(x, i) / fact(i + 1 ));
}
return total;
}
// Driver Code
public static void main(String[] args)
{
// Get x and n
int x = 5 , n = 4 ;
// Find and print the sum
System.out.print( "Sum is: " + sum(x, n));
}
} |
# Python3 Program to compute sum of # 1 + x / 2 ! + x ^ 2 / 3 ! +...+x ^ n/(n + 1)! # Method to find the factorial of a number def fact(n):
if n = = 1 :
return 1
else :
return n * fact(n - 1 )
# Method to compute the sum def sum (x, n):
total = 1.0
# Iterate the loop till n
# and compute the formula
for i in range ( 1 , n + 1 , 1 ):
total = total + ( pow (x, i) / fact(i + 1 ))
return total
# Driver code if __name__ = = '__main__' :
# Get x and n
x = 5
n = 4
# Print output
print ( "Sum is: {0:.4f}" . format ( sum (x, n)))
# This code is contributed by # SURENDRA_GANGWAR |
// C# Program to compute sum of // 1 + x/2! + x^2/3! +...+x^n/(n+1)! using System;
class SumOfSeries {
// Method to find factorial of a number
static int fact( int n)
{
if (n == 1)
return 1;
return n * fact(n - 1);
}
// Method to compute the sum
static double sum( int x, int n)
{
double total = 1.0;
// Iterate the loop till n
// and compute the formula
for ( int i = 1; i <= n; i++) {
total = total + (Math.Pow(x, i) / fact(i + 1));
}
return total;
}
// Driver Code
public static void Main()
{
// Get x and n
int x = 5, n = 4;
// Find and print the sum
Console.WriteLine( "Sum is: " + sum(x, n));
}
} // This code is contributed // by anuj_67.. |
<script> // java script Program to compute sum of // 1 + x/2! + x^2/3! +...+x^n/(n+1)! // Function to find the factorial // of a number function fact(n)
{ if (n == 1)
return 1;
return n * fact(n - 1);
} // Function to compute the sum function sum(x, n)
{ let total = 1.0;
// Iterate the loop till n
// and compute the formula
for (let i = 1; i <= n; i++)
{
total = total + (Math.pow(x, i) /
fact(i + 1));
}
return total.toFixed(4);
} // Driver code // Get x and n let x = 5; let n = 4; // Print output document.write( "Sum is: " + sum(x, n));
// This code is contributed by sravan kumar Gottumukkala </script> |
<?php // PHP Program to compute sum of // 1 + x/2! + x^2/3! +...+x^n/(n+1)! // Function to find the factorial // of a number function fact( $n )
{ if ( $n == 1)
return 1;
return $n * fact( $n - 1);
} // Function to compute the sum function sum( $x , $n )
{ $total = 1.0;
// Iterate the loop till n
// and compute the formula
for ( $i = 1; $i <= $n ; $i ++)
{
$total = $total + (pow( $x , $i ) /
fact( $i + 1));
}
return $total ;
} // Driver code // Get x and n $x = 5;
$n = 4;
// Print output echo "Sum is: " , sum( $x , $n );
// This code is contributed by ANKITRAI1 ?> |
Output
Sum is: 18.0833
Time Complexity: O(n2)
Auxiliary Space: O(n), since n extra space has been taken.
Efficient approach: Time complexity for above algorithm is O(
Below is the implementation of the above approach:
// C++ implementation of the approach #include <iostream> using namespace std;
// Function to compute the series sum double sum( int x, int n)
{ double total = 1.0;
// To store the value of S[i-1]
double previous = 1.0;
// Iterate over n to store sum in total
for ( int i = 1; i <= n; i++)
{
// Update previous with S[i]
previous = (previous * x) / (i + 1);
total = total + previous;
}
return total;
} // Driver code int main()
{ // Get x and n
int x = 5, n = 4;
// Find and print the sum
cout << "Sum is: " << sum(x, n);
return 0;
} // This code is contributed by jit_t |
// Java implementation of the approach public class GFG {
// Function to compute the series sum
static double sum( int x, int n)
{
double total = 1.0 ;
// To store the value of S[i-1]
double previous = 1.0 ;
// Iterate over n to store sum in total
for ( int i = 1 ; i <= n; i++) {
// Update previous with S[i]
previous = (previous * x) / (i + 1 );
total = total + previous;
}
return total;
}
// Driver code
public static void main(String[] args)
{
// Get x and n
int x = 5 , n = 4 ;
// Find and print the sum
System.out.print( "Sum is: " + sum(x, n));
}
} |
# Python implementation of the approach # Function to compute the series sum def sum (x, n):
total = 1.0 ;
# To store the value of S[i-1]
previous = 1.0 ;
# Iterate over n to store sum in total
for i in range ( 1 , n + 1 ):
# Update previous with S[i]
previous = (previous * x) / (i + 1 );
total = total + previous;
return total;
# Driver code if __name__ = = '__main__' :
# Get x and n
x = 5 ;
n = 4 ;
# Find and print the sum
print ( "Sum is: " , sum (x, n));
# This code is contributed by 29AjayKumar |
// C# implementation of the approach using System;
class GFG
{ // Function to compute the series sum
public double sum( int x, int n)
{
double total = 1.0;
// To store the value of S[i-1]
double previous = 1.0;
// Iterate over n to store sum in total
for ( int i = 1; i <= n; i++)
{
// Update previous with S[i]
previous = ((previous * x) / (i + 1));
total = total + previous;
}
return total;
}
} // Driver code class geek
{ public static void Main()
{
GFG g = new GFG();
// Get x and n
int x = 5, n = 4;
// Find and print the sum
Console.WriteLine( "Sum is: " + g.sum(x, n));
}
} // This code is contributed by SoM15242 |
<script> // Javascript implementation of the approach
// Function to compute the series sum
function sum(x, n)
{
let total = 1.0;
// To store the value of S[i-1]
let previous = 1.0;
// Iterate over n to store sum in total
for (let i = 1; i <= n; i++)
{
// Update previous with S[i]
previous = ((previous * x) / (i + 1));
total = total + previous;
}
return total;
}
// Get x and n
let x = 5, n = 4;
// Find and print the sum
document.write( "Sum is: " + sum(x, n));
</script> |
Output
Sum is: 18.083333333333336
Time Complexity: O(n)
Auxiliary Space: O(1) since using constant variables