Dado un número entero N, la tarea es encontrar la suma de la serie 2 0 + 2 1 + 2 2 + 2 3 + …. + 2 norte
Ejemplos:
Entrada: 5
Salida: 31
2 0 + 2 1 + 2 2 + 2 3 + 2 4
= 1 + 2+ 4 + 8 + 16
= 31
Entrada: 10
Salida: 1023
2 0 + 2 1 + 2 2 + 2 3 + 2 4 + 2 5 + 2 6 + 2 7 + 2 8 + 2 9
= 1 + 2+ 4 + 8 + 16 + 32 +64 + 128 + 256 + 512
= 1023
Un enfoque ingenuo es calcular la suma sumando cada potencia de 2 de 0 a n.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program to find sum
#include <bits/stdc++.h>
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 calculate 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)
PHP
<?php
// PHP 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
$sum = 0;
// loop to calculate
// sum of series
for ($i = 0; $i < $n; $i++)
{
// calculate 2^i
// and add it to sum
$sum = $sum + (1 << $i);
}
return $sum;
}
// Driver code
$n = 10;
echo "Sum of the series of " .
"power 2 is : ",
calculateSum($n);
// This code is contributed
// by Smitha
?>
Javascript
<script>
// 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
</script>
Producción:
Sum of series of power of 2 is : 1023
Complejidad de tiempo: O(n)
Un enfoque eficiente es encontrar el 2^(n+1) y restarle 1 ya que sabemos que 2^n se puede escribir como:
2n = ( 20+21+22+23+24 +...... 2n-1) +1
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program to find sum
#include <bits/stdc++.h>
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 calculate 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
}
PHP
<?php
// PHP program to calculate
// sum of series of power of 2
// function to calculate
// sum of series
function calculateSum($n)
{
// calculate 2^(n+1)
$sum = (1 << ($n + 1));
return $sum - 1;
}
// Driver code
$n = 10;
echo "Sum of the series of " .
"power 2 is : ",
calculateSum($n);
// This code is contributed
// by Smitha
Javascript
<script>
// 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
</script>
Producción:
Sum of series of power of 2 is :2047
Complejidad de tiempo: O(1)