系列3、5、9、17、33…的前n个项的总和。

给定n，我们需要找到表示为Sn = 3 + 5 + 9 + 17 + 33…的序列的前n个项的总和。
例子：

Input : 2
Output : 8
3 + 5 = 8

Input : 5
Output : 67
3 + 5 + 9 + 17 + 33 = 67

令，第n项由tn表示。
可以通过如下拆分每个术语来轻松解决此问题：

Sn = 3 + 5 + 9 + 17 + 33…… 
Sn = (2+1) + (4+1) + (8+1) + (16+1) +…… 
Sn = (2+1) + (2*2+1) + (2*2*2+1) + (2*2*2*2+1) +……+ ((2*2*2..unto n times) + 1)

我们观察到第n个项可以用2和1的幂表示。
因此，前n个项的总和如下：

Sn = (2+1) + (4+1) + (8+1) + (16+1) +……+ upto n terms 
Sn = (1 + 1 + 1 + 1 + …unto n terms) + (2 + 4 + 8 + 16 + …upto nth power of 2) 
In above formula, 
2 + 4 + 8 + 16…. is a G.P. 
It’s sum of first n terms is given by 2*(2^n-1)/(2-1) = 2^(n+1) – 2 (using G.P. formula) 
Sn = n + 2*(2^n – 1) 
Sn = 2^(n+1) + n -2

C++

// C++ program to find sum of first n terms
#include 
using namespace std;
 
int calculateSum(int n)
{
    // Sn = n*(4*n*n + 6*n - 1)/3
    return (pow(2, n + 1) + n - 2);
}
 
// Driver code
int main()
{
    // number of terms to be included in sum
    int n = 4;
 
    // find the Sn
    cout << "Sum = " << calculateSum(n);
 
    return 0;
}

Java

// Java program to find
// sum of first n terms
import java.util.*;
 
class GFG
{
static int calculateSum(int n)
{
    // Sn = n*(4*n*n + 6*n - 1)/3
    return ((int)Math.pow(2, n + 1) +
                             n - 2);
}
 
// Driver Code
public static void main(String args[])
{
    // number of terms to
    // be included in sum
    int n = 4;
 
    // find the Sn
    System.out.println("Sum = " +
                calculateSum(n));
}
}
 
// This code is contributed
// by Kirti_Mangal

Python

# Python program to find sum
# of n terms of the series
def calculateSum(n):
 
    return (2**(n + 1) + n - 2)
 
# Driver Code
 
# number of terms for the sum
n = 4
 
# find the Sn
print("Sum =", calculateSum(n))
 
# This code is contributed
# by Surendra_Gangwar

C#

//C# program to find
// sum of first n terms
using System;
 
class GFG
{
static int calculateSum(int n)
{
    // Sn = n*(4*n*n + 6*n - 1)/3
    return ((int)Math.Pow(2, n + 1) +
                            n - 2);
}
 
// Driver Code
public static void Main()
{
    // number of terms to
    // be included in sum
    int n = 4;
 
    // find the Sn
    Console.WriteLine("Sum = " +
                calculateSum(n));
}
}
 
// This code is contributed
// by inder_verma..

Javascript

// Java script program to find
// sum of first n terms
 
function calculateSum( n)
{
    // Sn = n*(4*n*n + 6*n - 1)/3
    return (Math.pow(2, n + 1) +
                             n - 2);
}
 
// Driver Code
 
    // number of terms to
    // be included in sum
    let n = 4;
 
    // find the Sn
    document.write("Sum = " +
                calculateSum(n));
 
// This code is contributed by mohan pavan
 

输出：

Sum = 34

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