 Open in App
Not now

# 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 termint calcSum(int n){    int a = pow(4, n);    return (4 * (a - 1) - 3 * n) / 9;} // Driver Codeint 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 approachimport 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 termdef calcSum(n):    a = pow(4, n)    return (4 * (a - 1) - 3 * n) / 9  # Driver Codeif __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 approachusing 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.

My Personal Notes arrow_drop_up