系列的总和2 ^ 0 + 2 ^ 1 + 2 ^ 2 +….. + 2 ^ n

给定一个整数N，任务是找到2 ⁰ + 2 ¹ + 2 ² + 2 ³ +…的和。 + 2 ⁿ 。
例子：

Input: 5 
Output: 31 
2⁰ + 2¹ + 2² + 2³ + 2⁴ 
= 1 + 2+ 4 + 8 + 16 
= 31
Input: 10 
Output: 1023 
2⁰ + 2¹ + 2 ² + 2³ + 2 ⁴ + 2⁵ + 2⁶ + 2⁷ + 2 ⁸ + 2⁹ 
= 1 + 2+ 4 + 8 + 16 + 32 +64 + 128 + 256 + 512 
= 1023

一个幼稚的方法是计算总和是将2的每个幂乘以0到n。
下面是上述方法的实现：

C++

// C++ program to find sum
#include 
using namespace std;
 
// function to calculate sum of series
int calculateSum(int n)
{
    // initialize sum as 0
    int sum = 0;
 
    // loop to calculate sum of series
    for (int i = 0; i < n; i++) {
 
        // calculate 2^i
        // and add it to sum
 
        sum = sum + (1 << i);
    }
    return sum;
}
 
// Driver code
int main()
{
    int n = 10;
    cout << "Sum of series of power of 2 is : "
         << calculateSum(n);
}

Java

// Java program to find sum
class GFG {
    // function to calculate sum of series
    static int claculate sum(int n)
    {
        // initialize sum as 0
        int sum = 0;
 
        // loop to calculate sum of series
        for (int i = 0; i < n; i++) {
 
            // calculate 2^i
            // and add it to sum
 
            sum = sum + (1 << i);
        }
        return sum;
    }
    // Main function
    public static void main(String[] args)
    {
 
        int n = 10;
        System.out.println("Sum of the series : " + calculateSum(n));
    }
};

Python3

# Python3 program to calculate
# sum of series of power of 2
 
# function to calculate sum of series
def calculateSum(n):
     sum = 0
 
     # loop to calculate sum of series
     for i in range (0, n):
 
         # calculate 2 ^ i
         sum = sum+ (1 << i)
    
     return sum
 
# Driver code
n = 10
print("Sum of series ", calculateSum(n))

C#

// C# program to find sum
using System;
 
class GFG
{
    // function to calculate
    // sum of series
    static int calculateSum(int n)
    {
        // initialize sum as 0
        int sum = 0;
 
        // loop to calculate
        // sum of series
        for (int i = 0; i < n; i++)
        {
 
            // calculate 2^i
            // and add it to sum
 
            sum = sum + (1 << i);
        }
        return sum;
    }
     
    // Driver code
    public static void Main()
    {
        int n = 10;
        Console.WriteLine("Sum of the series : " +
                                 calculateSum(n));
    }
}
 
// This code is contributed
// by Akanksha Rai(Abby_akku)

Javascript

 
// Javascript  program to find sum of the
// series 2^0 + 2^1 + 2^2 +…..+ 2^n
 
// function to calculate
// sum of series
function calculateSum(n)
{
    // initialize sum as 0
    let sum = 0;
 
    // loop to calculate
    // sum of series
    for (let i = 0; i < n; i++)
    {
 
        // calculate 2^i
        // and add it to sum
 
        sum = sum + (1 << i);
    }
    return sum;
}
 
// Driver code
let n = 10;
document.write("Sum of the series of power 2 is : "
                           + calculateSum(n))
 
//This code is contributed by sravan kumar

---

输出：

Sum of series of power of 2 is : 1023

时间复杂度： O(n)
一种有效的方法是找到2 ^(n + 1)并从中减去1，因为我们知道2 ^ n可以写成：

2n = ( 20+21+22+23+24 +...... 2n-1) +1

下面是上述方法的实现：

C++

// C++ program to find sum
#include 
using namespace std;
 
int calculateSum(int n)
{
    // calculate  and return 2^(n+1) -1
    return (1 << (n + 1)) - 1;
}
 
int main()
{
    int n = 10;
    cout << "Sum of series of power of 2 is :"
         << calculateSum(n);
}

Java

// Java program to calculate
// sum of series of power of 2
 
class GFG {
 
    // function to calculate sum of series
    static int claculate sum(int n)
    {
 
        // calculate 2^(n+1)
        int sum = (1 << (n + 1));
        return sum - 1;
    }
 
    // Driver code
    public static void main(String[] args)
    {
 
        int n = 10;
        System.out.println("Sum of the series of power 2 is : "
                           + calculateSum(n));
    }
};

Python3

# Python3 program to calculate
# sum of series of 2's power
 
# function to calculate sum of series
def calculateSum(n):
 
      # calculate 2^(n + 1)
      sum = (1 << (n + 1))
      return sum-1
 
# Driver code
n = 10
print("Sum of series ", calculateSum(n))

C＃

// C# program to calculate
// sum of series of power of 2
using System;
class GFG
{
 
    // function to calculate
    // sum of series
    static int calculateSum(int n)
    {
 
        // calculate 2^(n+1)
        int sum = (1 << (n + 1));
        return sum - 1;
    }
 
    // Driver code
    public static void Main()
    {
        int n = 10;
        Console.Write("Sum of the series " +
                        "of power 2 is : " +
                           calculateSum(n));
    }
     
// This code is contributed
// by Smitha
}

Java脚本

 
// Javascript program to calculate
// sum of series of power of 2
 
// function to calculate
// sum of series
function calculateSum(n)
{
 
    // calculate 2^(n+1)
    let sum = (1 << (n + 1));
    return sum - 1;
}
 
// Driver code
let n = 10;
document.write("Sum of the series of power 2 is : "
                           + calculateSum(n));
 
// This code is contributed by sravan kumar
 

输出：

Sum of series of power of 2 is :2047

时间复杂度： O(1)