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# Find sum of N terms of series 1, (1+4) , (1+4+4^2), (1+4+4^2+4^3), …..

• Last Updated : 20 Aug, 2022

Given a positive integer, N. Find the sum of the first N term of the series-

1, (1+4), (1+4+42), (1+4+42+43), …., till N terms

Examples:

Input: N = 3
Output: 27

Input: N = 5
Output: 453

Approach:

1st term = 1

2nd term = (1 + 4)

3rd term = (1 + 4 + 4 ^ 2)

4th term = (1 + 4 + 4 ^ 2 + 4 ^ 3)

.

.

Nth term = (1 + 4 + 4 ^ 2+….+ 4 ^ (N – 2) + 4 ^(N – 1))

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

The following series of steps can be used to derive the formula to find the sum of N terms-

The series can be decomposed as-     -(1)

The equation (1) is in G.P. with

First term a = 1

Common ration r = 4

The sum of N terms in G.P. for r>1 is Substituting the values of a and r in the above equation, we get- Thus, the term The sum of the series 1, (1+4), (1+4+4^{2}), (1+4+4^{2}+4^{3})+….+N terms can be represented as-    -(2)

The equation- is in G.P. with

First term a = 4

Common ratio r = 4

Applying the formula of sum of G.P.- -(3)

Substituting equation (3) in equation (2), we get-   Illustration:

Input: N = 3
Output: 11
Explanation:    Below is the implementation of the above approach:

## C++

 // C++ program to implement // the above approach #include  using namespace std;   // Function to calculate the sum // of first N term int calcSum(int n) {     int a = pow(4, n);     return (4 * (a - 1) - 3 * n) / 9; }   // Driver Code int main() {     // Value of N     int N = 3;       // Function call to calculate     // sum of the series     cout << calcSum(N);     return 0; }

## Java

 // Java code for the above approach import java.util.*;   class GFG{     // Function to calculate the sum   // of first N term   static int calcSum(int n)   {     int a = (int)Math.pow(4, n);     return (4 * (a - 1) - 3 * n) / 9;   }       // Driver Code   public static void main(String[] args)   {     // Value of N     int N = 3;       // Function call to calculate     // sum of the series     System.out.print(calcSum(N));   } }   // This code is contributed by code_hunt.

## Python3

 # Python 3 program for the above approach   # Function to calculate the sum # of first N term def calcSum(n):     a = pow(4, n)     return (4 * (a - 1) - 3 * n) / 9     # Driver Code if __name__ == "__main__":       # Value of N     N = 3           # Function call to calculate     # sum of the series     print(calcSum(N))   # This code is contributed by Abhishek Thakur.

## C#

 // C# code for the above approach using System;   class GFG{     // Function to calculate the sum   // of first N term   static int calcSum(int n)   {     int a = (int)Math.Pow(4, n);     return (4 * (a - 1) - 3 * n) / 9;   }       // Driver Code   public static void Main()   {     // Value of N     int N = 3;       // Function call to calculate     // sum of the series     Console.Write(calcSum(N));   } }   // This code is contributed by gfgking

## Javascript

 

Output

27

Time Complexity: O(log4n) because using inbuilt pow function
Auxiliary Space: O(1), since no extra space has been taken.

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