Sum of series 2/3 – 4/5 + 6/7 – 8/9 + ——- upto n terms
Given the value of n, find the sum of the series (2 / 3) – (4 / 5) + (6 / 7) – (8 / 9) + – – – – – – – upto n terms.
Examples :
Input : n = 5
Output : 0.744012
Series : (2 / 3) - (4 / 5) + (6 / 7) - (8 / 9) + (10 / 11)
Input : n = 7
Output : 0.754268
Series : (2 / 3) - (4 / 5) + (6 / 7) - (8 / 9) +
(10 / 11) - (12 / 13) + (14 / 15)
C++
#include <bits/stdc++.h>
using namespace std;
double seriesSum( int n)
{
int i = 1;
double res = 0.0;
bool sign = true ;
while (n > 0)
{
n--;
if (sign) {
sign = !sign;
res = res + ( double )++i / ++i;
}
else {
sign = !sign;
res = res - ( double )++i / ++i;
}
}
return res;
}
int main()
{
int n = 5;
cout << seriesSum(n);
return 0;
}
|
Java
import java.io.*;
class GFG {
static double seriesSum( int n)
{
int i = 1 ;
double res = 0.0 ;
boolean sign = true ;
while (n > 0 )
{
n--;
if (sign)
{
sign = !sign;
res = res + ( double )++i / ++i;
}
else
{
sign = !sign;
res = res - ( double )++i / ++i;
}
}
return res;
}
public static void main (String[] args) {
int n = 5 ;
System.out.print(seriesSum(n));
}
}
|
Python3
def seriesSum(n):
i = 1 ;
res = 0.0 ;
sign = True ;
while (n > 0 ):
n = n - 1 ;
if (sign):
sign = False ;
res = res + (i + 1 ) / (i + 2 );
i = i + 2 ;
else :
sign = True ;
res = res - (i + 1 ) / (i + 2 );
i = i + 2 ;
return res;
n = 5 ;
print ( round (seriesSum(n), 6 ));
|
C#
using System;
class GFG {
static double seriesSum( int n)
{
int i = 1;
double res = 0.0;
bool sign = true ;
while (n > 0)
{
n--;
if (sign)
{
sign = !sign;
res = res + ( double )++i / ++i;
}
else
{
sign = !sign;
res = res - ( double )++i / ++i;
}
}
return res;
}
public static void Main () {
int n = 5;
Console.Write(seriesSum(n));
}
}
|
PHP
<?php
function seriesSum( $n )
{
$i = 1;
$res = 0.0;
$sign = true;
while ( $n > 0)
{
$n --;
if ( $sign ) {
$sign = ! $sign ;
$res = $res + (double)++ $i / ++ $i ;
}
else {
$sign = ! $sign ;
$res = $res - (double)++ $i / ++ $i ;
}
}
return $res ;
}
$n = 5;
echo (seriesSum( $n ));
?>
|
Javascript
<script>
function seriesSum( n)
{
let i = 1;
let res = 0.0;
let sign = true ;
while (n > 0)
{
n--;
if (sign) {
sign = !sign;
res = res + ++i / ++i;
}
else {
sign = !sign;
res = res - ++i / ++i;
}
}
return res;
}
let n = 5 ;
document.write(seriesSum(n).toFixed(6)) ;
</script>
|
Output :
0.744012
Time Complexity: O(n), where n represents the given integer.
Auxiliary Space: O(1), no extra space is required, so it is a constant.
Last Updated :
20 Feb, 2023
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Sum of the series 1, 2, 4, 3, 5, 7, 9, 6, 8, 10, 11, 13.. till N-th term
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