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

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:

27

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