求系列 1, 4, 13, 40, 121, ... 的 N 项之和

给定一个正整数n 。求系列的前n 项之和：

1, 4, 13, 40, 121, …..

例子：

Input: n = 5
Output: 179

Input: n = 3
Output: 18

方法：

该序列是通过使用以下模式形成的。对于任何值 N-

$S_{n}=(\frac{3^{n+1}-3-2*n}{4})$

上述解决方案可以通过以下一系列步骤得出 -

Let T_n be the nth term and S_n be the sum to n terms of the given series.

Thus, we have

$S_{n}=1+4+13+40+121+...+T_{n-1}+T_{n}$  -(1)

Equation (1) can be written as-

S_{n}=1+4+13+40+121+…+T_{n-1}+T_{n}  -(2)

Subtracting equation (2) from equation (1), we get

$0=1+[3+9+27+81+...+(T_{n}-T_{n-1})-T_{n}]$

The above equation 3, 9, 27, 81, … is a G.P. with common ratio 3 and first term 3.

Thus, we have

$1+[\frac{3*(3^{n-1}-1)}{3-1}]-T_{n}=0$

$1+[\frac{3*(3^{n-1}-1)}{2}]-T_{n}=0$

$(\frac{3^{n}}{2}-\frac{1}{2})-T_{n}=0$

$(\frac{3^{n}}{2}-\frac{1}{2})=T_{n}$

Since,

$S_{n}=\sum_{k=1}^{n}T_{k}$

Therefore,

$S_{n}=\sum_{k=1}^{n}(\frac{3^{k}}{2}-\frac{1}{2})$

$S_{n}=\frac{1}{2}\sum_{k=1}^{n}3^{k}-\frac{1}{2}\sum_{k=1}^{n}1$

$S_{n}=\frac{1}{2}(3+3^{2}+3^{3}+3^{4}+3^{5}+...+3^{n})-\frac{n}{2}$

$S_{n}=\frac{1}{2}[\frac{3*(3^{n}-1)}{2}]-\frac{n}{2}$

$S_{n}=(\frac{3^{n+1}-3}{4})-\frac{n}{2}$

Thus, sum of n terms is

$S_{n}=(\frac{3^{n+1}-3-2*n}{4})$

插图：

Input: n = 5
Output: 179
Explanation:
$S_{n}=(\frac{3^{n+1}-3-2*n}{4})$
       =$(\frac{3^{5+1}-3-2*5}{4})$
       =$(\frac{716}{4})$
       =179

下面是上述方法的实现：

C++

// C++ program to implement
// the above approach
#include 
using namespace std;
 
// Function to return sum of
// first N term of the series
int findSum(int N)
{
    return (pow(3, N + 1) -
                3 - 2 * N) / 4;
}
 
// Driver Code
int main()
{
    int N = 5;
    cout << findSum(N);
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
public class GFG
{
 
  // Function to return sum of
  // first N term of the series
  static int findSum(int N)
  {
    return (int)(Math.pow(3, N + 1) -
                 3 - 2 * N) / 4;
  }
 
  public static void main(String args[])
  {
    int N = 5;
    System.out.print(findSum(N));
  }
}
// This code is contributed by Samim Hossain Mondal.

Python3

# Python code for the above approach
 
# Function to return sum of
# first N term of the series
def findSum(N):
    return (3 ** (N + 1) - 3 - 2 * N) // 4;
 
# Driver Code
N = 5;
print(findSum(N));
 
# This code is contributed by Saurabh Jaiswal

C#

// C# program to implement
// the above approach
using System;
class GFG
{
 
// Function to return sum of
// first N term of the series
static int findSum(int N)
{
    return (int)(Math.Pow(3, N + 1) -
                3 - 2 * N) / 4;
}
 
  // Driver Code
  public static void Main()
  {
    int N = 5;
    Console.Write(findSum(N));
 
  }
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript

      // JavaScript code for the above approach
 
      // Function to return sum of
      // first N term of the series
      function findSum(N) {
          return Math.floor((Math.pow(3, N + 1) -
              3 - 2 * N) / 4);
      }
 
      // Driver Code
      let N = 5;
      document.write(findSum(N));
 
// This code is contributed by Potta Lokesh
  

输出

179

时间复杂度： O(1)

辅助空间： O(1)