This is a mathematical series program where a user must enter the number of terms up to which the sum of the series is to be found. Following this, we also need the value of x, which forms the base of the series.
Examples :
Input : base = 2, range = 5
Output : 18.07
Input : base = 1, range = 10
Output : 3.93
Method 1 (Simple) We just need to follow the series and put the values of the base at x and value range at n and get the sum.
C++
#include <math.h>
#include <iostream>
#include <boost/format.hpp>
class gfg
{
public :
double sum( int x, int n)
{
double i, total = 1.0;
for (i = 1; i <= n; i++)
total = total +
( pow (x, i) / i);
return total;
}
};
int main()
{
gfg g;
int x = 2;
int n = 5;
std::cout << boost::format( "%.2f" ) % g.sum(x,n);
return 0;
}
|
C
#include <math.h>
#include <stdio.h>
double sum( int x, int n)
{
double i, total = 1.0;
for (i = 1; i <= n; i++)
total = total +
( pow (x, i) / i);
return total;
}
int main()
{
int x = 2;
int n = 5;
printf ( "%.2f" , sum(x, n));
return 0;
}
|
Java
import static java.lang.Math.pow;
class GFG
{
static double sum( int x, int n)
{
double i, total = 1.0 ;
for (i = 1 ; i <= n; i++)
total = total +
(Math.pow(x, i) / i);
return total;
}
public static void main(String[] args)
{
int x = 2 ;
int n = 5 ;
System.out.printf( "%.2f" , sum(x, n));
}
}
|
Python3
def SUM (x, n):
total = 1
for i in range ( 1 , n + 1 ):
total = total + ((x * * i) / i)
return total
x = 2
n = 5
s = SUM (x, n)
print ( round (s, 2 ))
|
C#
using System;
class GFG
{
static float sum( int x, int n)
{
double i, total = 1.0;
for (i = 1; i <= n; i++)
total = total +
(Math.Pow(x, i) / i);
return ( float )total;
}
public static void Main()
{
int x = 2;
int n = 5;
Console.WriteLine(sum(x, n));
}
}
|
PHP
<?php
function sum( $x , $n )
{
$i ; $total = 1.0;
for ( $i = 1; $i <= $n ; $i ++)
$total = $total +
(pow( $x , $i ) / $i );
return $total ;
}
$x = 2;
$n = 5;
echo (sum( $x , $n ));
?>
|
Javascript
<script>
function sum(x, n)
{
let i, total = 1.0;
for (i = 1; i <= n; i++)
total = total +
(Math.pow(x, i) / i);
return total;
}
let g;
let x = 2;
let n = 5;
document.write(sum(x, n).toFixed(2));
</script>
|
Output :
18.07
Time Complexity: O(nlogn)
Auxiliary Space: O(1), since no extra space has been taken>
Method 2 (Optimized) We can avoid use of pow() function and reuse the previously computed power.
C++
#include <bits/stdc++.h>
using namespace std;
double sum( int x, int n)
{
double i, total = 1.0, multi = x;
for (i = 1; i <= n; i++)
{
total = total + multi / i;
multi = multi * x;
}
return total;
}
int main()
{
int x = 2;
int n = 5;
cout << fixed << setprecision(2) << sum(x, n);
return 0;
}
|
C
#include <math.h>
#include <stdio.h>
double sum( int x, int n)
{
double i, total = 1.0, multi = x;
for (i = 1; i <= n; i++) {
total = total + multi / i;
multi = multi * x;
}
return total;
}
int main()
{
int x = 2;
int n = 5;
printf ( "%.2f" , sum(x, n));
return 0;
}
|
Java
class GFG
{
static double sum( int x, int n)
{
double i, total = 1.0 , multi = x;
for (i = 1 ; i <= n; i++)
{
total = total + multi / i;
multi = multi * x;
}
return total;
}
public static void main(String[] args)
{
int x = 2 ;
int n = 5 ;
System.out.printf( "%.2f" , sum(x, n));
}
}
|
Python3
def sum (x, n):
total = 1.0
multi = x
for i in range ( 1 , n + 1 ):
total = total + multi / i
multi = multi * x
return total
x = 2
n = 5
print ( round ( sum (x, n), 2 ))
|
C#
using System;
class GFG
{
static float sum( int x, int n)
{
double i, total = 1.0, multi = x;
for (i = 1; i <= n; i++)
{
total = total + multi / i;
multi = multi * x;
}
return ( float )total;
}
public static void Main()
{
int x = 2;
int n = 5;
Console.WriteLine(sum(x, n));
}
}
|
PHP
<?php
function sum( $x , $n )
{
$i ; $total = 1.0; $multi = $x ;
for ( $i = 1; $i <= $n ; $i ++)
{
$total = $total + $multi / $i ;
$multi = $multi * $x ;
}
return $total ;
}
$x = 2;
$n = 5;
echo (sum( $x , $n ));
?>
|
Javascript
<script>
function sum(x, n)
{
let total = 1.0;
let multi = x;
for (let i = 1; i <= n; i++)
{
total = total + multi / i;
multi = multi * x;
}
return total;
}
let x = 2;
let n = 5;
document.write(sum(x, n).toFixed(2));
</script>
|
Output :
18.07
Time complexity: O(n) since using a for loop
Auxiliary Space: O(1), since no extra space has been taken.
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