Find the sum of series 3, -6, 12, -24 . . . upto N terms

Given an integer N. The task is to find the sum upto N terms of the given series:

3, -6, 12, -24, … upto N terms

Examples: 

Input : N = 5
Output : Sum = 33

Input : N = 20
Output : Sum = -1048575

On observing the given series, it can be seen that the ratio of every term with their previous term is same which is -2. Hence the given series is a GP(Geometric Progression) series.
You can learn more about GP series here.
So, when r < 0.
In above GP series the first term i:e a = 3 and common ratio i:e r = (-2).
Therefore, . 
Thus, .
Below is the implementation of above approach: 

C++

//C++ program to find sum upto N term of the series:
// 3, -6, 12, -24, .....

#include<iostream>
#include<math.h>
using namespace std;
//calculate sum upto N term of series

class gfg
{
    public:
    int Sum_upto_nth_Term(int n)
    {
        return (1 - pow(-2, n));
    }
};
// Driver code
int main()
{
    gfg g;
    int N = 5;
    cout<<g.Sum_upto_nth_Term(N);
}

Java

//Java program to find sum upto N term of the series:
// 3, -6, 12, -24, .....

import java.util.*;
//calculate sum upto N term of series

class solution
{

static int Sum_upto_nth_Term(int n)
{
    return (1 -(int)Math.pow(-2, n));
}

// Driver code
public static void main (String arr[])
{
    int N = 5;
    System.out.println(Sum_upto_nth_Term(N));
}

}

Python

# Python program to find sum upto N term of the series:
# 3, -6, 12, -24, .....

# calculate sum upto N term of series
def Sum_upto_nth_Term(n):
    return (1 - pow(-2, n))

# Driver code
N = 5
print(Sum_upto_nth_Term(N))

C#

// C# program to find sum upto
// N term of the series:
// 3, -6, 12, -24, .....

// calculate sum upto N term of series
class GFG
{

static int Sum_upto_nth_Term(int n)
{
    return (1 -(int)System.Math.Pow(-2, n));
}

// Driver code
public static void Main()
{
    int N = 5;
    System.Console.WriteLine(Sum_upto_nth_Term(N));
}
}

// This Code is contributed by mits

PHP

<?php
// PHP program to find sum upto
// Nth term of the series:
// 3, -6, 12, -24, .....

// calculate sum upto N term of series
function Sum_upto_nth_Term($n)
{
    return (1 - pow(-2, $n));
}

// Driver code
$N = 5;
echo (Sum_upto_nth_Term($N));

// This code is contributed
// by Sach_Code
?>

Javascript

<script>
// Java program to find sum upto N term of the series:
// 3, -6, 12, -24, .....

// calculate sum upto N term of series
function Sum_upto_nth_Term( n) {
    return (1 - parseInt( Math.pow(-2, n)));
}

// Driver code

    let N = 5;
    document.write(Sum_upto_nth_Term(N));

// This code is contributed by 29AjayKumar
</script>

Output:

33

Time Complexity: O(logn), where n is the given integer.

Auxiliary Space: O(1), no extra space is required, so it is a constant.