Sum of the series 1 / 1 + (1 + 2) / (1 * 2) + (1 + 2 + 3) / (1 * 2 * 3) + – – – – – – upto n terms.
Given a positive integer n, find the sum of the series upto n terms.
Examples :
Input : n = 4
Output : 3.91667
Series : (1 / 1) + (1 + 2) / (1 * 2) + (1 + 2 + 3) / (1 * 2 * 3) + (1 + 2 + 3 + 4) / (1 * 2 * 3 * 4)
= 1 / 1 + 3 / 2 + 6 / 6 + 10 / 24
= 3.91667
Input : n = 6
Output : 4.07083
Series : (1 / 1) + (1 + 2) / (1 * 2) + (1 + 2 + 3) / (1 * 2 * 3) + (1 + 2 + 3 + 4) / (1 * 2 * 3 * 4) + (1 + 2 + 3 + 4 + 5) / (1 * 2 * 3 * 4 * 5) + (1 + 2 + 3 + 4 + 5 + 6) / (1 * 2 * 3 * 4 * 5 * 6)
= 1 / 1 + 3 / 2 + 6 / 6 + 10 / 24 + 15 / 120 + 21 / 720
= 4.07083
C++
#include <bits/stdc++.h>
using namespace std;
double sumOfSeries( int n)
{
double res = 0.0 ;
int sum = 0, prod = 1;
for ( int i = 1 ; i <= n ; i++)
{
sum += i;
prod *= i;
res += (( double )sum / prod);
}
return res;
}
int main()
{
int n = 4 ;
cout << sumOfSeries(n) ;
return 0;
}
|
Java
class GFG
{
static double sumOfSeries( int n)
{
double res = 0.0 ;
int sum = 0 , prod = 1 ;
for ( int i = 1 ; i <= n; i++) {
sum += i;
prod *= i;
res += (( double )sum / prod);
}
return res;
}
public static void main(String arg[]) {
int n = 4 ;
System.out.println(sumOfSeries(n));
}
}
|
Python3
def sumOfSeries(n) :
res = 0.0
sum = 0
prod = 1
for i in range ( 1 , n + 1 ) :
sum = sum + i
prod = prod * i
res = res + ( sum / prod)
return res
n = 4
print ( round (sumOfSeries(n), 5 ))
|
C#
using System;
class GFG {
static double sumOfSeries( int n)
{
double res = 0.0;
int sum = 0, prod = 1;
for ( int i = 1; i <= n; i++) {
sum += i;
prod *= i;
res += (( double )sum / prod);
}
return res;
}
public static void Main()
{
int n = 4;
Console.Write(sumOfSeries(n));
}
}
|
PHP
<?php
function sumOfSeries( $n )
{
$res = 0.0 ;
$sum = 0; $prod = 1;
for ( $i = 1 ; $i <= $n ; $i ++)
{
$sum += $i ;
$prod *= $i ;
$res += ((double) $sum / $prod );
}
return $res ;
}
$n = 4 ;
echo (sumOfSeries( $n )) ;
?>
|
Javascript
<script>
function sumOfSeries(n)
{
let res = 0.0 ;
let sum = 0, prod = 1;
for (let i = 1 ; i <= n ; i++)
{
sum += i;
prod *= i;
res += (sum / prod);
}
return res;
}
let n = 4 ;
document.write(sumOfSeries(n).toFixed(5)) ;
</script>
|
Output :
3.91667
Time complexity: O(n) since using a single loop
Auxiliary Space: O(1) for constant space for variables
Last Updated :
20 Feb, 2023
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