Find the sum of N terms of the series 1, 4, 13, 40, 121, …

Given a positive integer, n. Find the sum of the first n term of the series:

1, 4, 13, 40, 121, …..

Examples:

Input: n = 5
Output: 179

Input: n = 3
Output: 18

Approach:

The sequence is formed by using the following pattern. For any value N-

The above solution can be derived following the series of steps-

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

Thus, we have

  -(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

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

Thus, we have

Since,

Therefore,

Thus, sum of n terms is

Illustration:

Input: n = 5
Output: 179
Explanation:

=
       =
       =179

Below is the implementation of the above approach:

C++

// C++ program to implement
// the above approach
#include <bits/stdc++.h>
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

<script>
      // 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
  </script>

Output

179

Time Complexity: O(log₃N) because using inbuilt pow function

Auxiliary Space: O(1)