求出系列2、5、13、35、97的总和…

给定一个序列和一个数字n，任务是找到其前n个项的总和。以下是系列：

2, 5, 13, 35, 97, …

例子：

Input: N = 2
Output: 7
The sum of first 2 terms of Series is
2 + 5 = 7

Input: N = 4
Output: 55
The sum of first 4 terms of Series is
2 + 5 + 13 + 35 = 55

方法：从这个给定的序列中，我们发现它是两个具有公共比率2，3的GP系列的和。

Sn = 2 + 5 + 13 + 35 + 97 … + upto nth term 
Sn = (2^0 + 3^ 0) + (2^1 + 3^1) + (2^2 + 3^2) + (2^3 + 3^3)+ (2^4 + 3^4) …… + upto nth term 
Sn = (2^0 + 2^1 + 2^2 + 2^3 + 2^4 … + upto nth term) + ( 3^0 + 3^1 + 3^2 + 3^3 …… + unto nth term )

因为，我们知道GP的n个项的总和由以下公式给出：

    $$$ S_n=\frac{a(r^n-1)}{(r-1)} $$$

下面是上述方法的实现：

C++

// C++ program for finding the sum
// of first N terms of the series.
#include 
using namespace std;
 
// CalculateSum function returns the final sum
int calculateSum(int n)
{
    // r1 and r2 are common ratios
    // of 1st and 2nd series
    int r1 = 2, r2 = 3;
 
    // a1 and a2 are common first terms
    // of 1st and 2nd series
    int a1 = 1, a2 = 1;
 
    return a1 * (pow(r1, n) - 1) / (r1 - 1)
           + a2 * (pow(r2, n) - 1) / (r2 - 1);
}
 
// Driver code
int main()
{
    int n = 4;
 
    // function calling for 4 terms
    cout << "Sum = " << calculateSum(n)
         << endl;
 
    return 0;
}

Java

//Java program for finding the sum
//of first N terms of the series.
 
public class GFG {
 
    //CalculateSum function returns the final sum
    static int calculateSum(int n)
    {
     // r1 and r2 are common ratios
     // of 1st and 2nd series
     int r1 = 2, r2 = 3;
 
     // a1 and a2 are common first terms
     // of 1st and 2nd series
     int a1 = 1, a2 = 1;
 
     return (int)(a1 * (Math.pow(r1, n) - 1) / (r1 - 1)
            + a2 * (Math.pow(r2, n) - 1) / (r2 - 1));
    }
 
    //Driver code
    public static void main(String[] args) {
         
        int n = 4;
 
         // function calling for 4 terms
         System.out.println("Sum = " +calculateSum(n));
    }
}

Python 3

# Python 3 program for finding the sum
# of first N terms of the series.
 
# from math import everything
from math import *
 
# CalculateSum function returns the final sum
def calculateSum(n) :
 
    # r1 and r2 are common ratios
    # of 1st and 2nd series
    r1, r2 = 2, 3
 
    # a1 and a2 are common first terms
    # of 1st and 2nd series
    a1, a2 = 1, 1
 
    return (a1 * (pow(r1, n) - 1) // (r1 - 1)
           + a2 * (pow(r2, n) - 1) // (r2 - 1))
 
# Driver Code
if __name__ == "__main__" :
 
    n = 4
 
    # function calling for 4 terms
    print("SUM = ",calculateSum(n))
 
 
# This code is contributed by ANKITRAI1

C#

// C# program for finding the sum
// of first N terms of the series.
using System;
 
class GFG
{
 
// CalculateSum function
// returns the final sum
static int calculateSum(int n)
{
// r1 and r2 are common ratios
// of 1st and 2nd series
int r1 = 2, r2 = 3;
 
// a1 and a2 are common first
// terms of 1st and 2nd series
int a1 = 1, a2 = 1;
 
return (int)(a1 * (Math.Pow(r1, n) - 1) / (r1 - 1) +
             a2 * (Math.Pow(r2, n) - 1) / (r2 - 1));
}
 
// Driver code
static public void Main ()
{
    int n = 4;
 
    // function calling for 4 terms
    Console.Write("Sum = " +
                   calculateSum(n));
}
}
 
// This code is contributed by Raj

Javascript

 
//javascript program for finding the sum
//of first N terms of the series.
//CalculateSum function returns the final sum
function calculateSum(n)
{
 // r1 and r2 are common ratios
 // of 1st and 2nd series
 var r1 = 2, r2 = 3;
 
 // a1 and a2 are common first terms
 // of 1st and 2nd series
 var a1 = 1, a2 = 1;
 
 return parseInt((a1 * (Math.pow(r1, n) - 1) / (r1 - 1)
        + a2 * (Math.pow(r2, n) - 1) / (r2 - 1)));
}
 
    //Driver code
     
         
var n = 4;
 
 // function calling for 4 terms
 document.write("Sum = " +calculateSum(n));
 
// This code contributed by shikhasingrajput

输出：

Sum = 55