# First term from given Nth term of the equation F(N) = (2 * F(N – 1)) % 10^9 + 7

Given an integer N and an integer FN which denotes the Nth term of the linear equation F(N) = (2 * F(N – 1)) % M, where M is 109 + 7, the task is to find the value of F(1).

Examples:

Input : N = 2, FN = 6
Output:
Explanation:
If F(1) = 3, F(2) = (2 * F(1)) % M = (2 * 3) % M = 6.
For F(1) = 3 the given linear equation satisfies the value of F(2).
Therefore, the value of F(1) is 3.

Input : N = 3, FN = 6
Output: 500000005
Explanation:
If F(1) = 500000005
F(2) = (2 * 500000005) % M = 3
F(3) = (2 * 3) % M = 6
For F(1) = 500000005 the given linear equation satisfies the value of F(3).
Therefore, the value of F(1) is 500000005

Naive Approach: The simplest approach to solve this problem is to try all possible values of F(1) in the range [1, M – 1] and check if any value satisfies the given linear equation or not. If found to be true, then print the value of F(1).

Time Complexity: O(N * M)
Auxiliary Space: O(1)

Efficient Approach: To optimize the above approach the idea is based on the following observations:

Given the linear equation, F(N) = 2 * F(N – 1) ——(1)
put the value of F(N – 1) = 2 * F(N – 2) in equation(1)
=>F(N) = 2 * (2 * F(N – 2)) ——–(2)
put the value of F(N – 2) = 2 * F(N – 3) in equation(2)
=>F(N) = 2* (2 * (2 * F(N – 3)))
…………………………….
…………………………….
=>F(N) = 2(N – 1) F(N – (N – 1)) = 2(N – 1) F(1)
=>F(1) = F(N) / 2(N – 1)

Follow the steps below to solve the problem:

Below is the implementation of the above approach:

## C++

 `// C++ program to implement` `// the above approach`   `#include ` `using` `namespace` `std;` `#define M 1000000007`   `// Function to find the value` `// of power(X, N) % M` `long` `long` `power(``long` `long` `x,` `                ``long` `long` `N)` `{` `    ``// Stores the value` `    ``// of (X ^ N) % M` `    ``long` `long` `res = 1;`   `    ``// Calculate the value of` `    ``// power(x, N) % M` `    ``while` `(N > 0) {`   `        ``// If N is odd` `        ``if` `(N & 1) {`   `            ``// Update res` `            ``res = (res * x) % M;` `        ``}`   `        ``// Update x` `        ``x = (x * x) % M;`   `        ``// Update N` `        ``N = N >> 1;` `    ``}` `    ``return` `res;` `}`   `// Function to find modulo multiplicative` `// inverse of X under modulo M` `long` `long` `moduloInverse(``long` `long` `X)` `{` `    ``return` `power(X, M - 2);` `}`   `// Function to find the value of F(1)` `long` `long` `F_1(``long` `long` `N,` `              ``long` `long` `F_N)` `{`   `    ``// Stores power(2, N - 1)` `    ``long` `long` `P_2 = power(2, N - 1);`   `    ``// Stores modulo multiplicative` `    ``// inverse of P_2 under modulo M` `    ``long` `long` `modInv = moduloInverse(P_2);`   `    ``// Stores the value of F(1)` `    ``long` `long` `res;`   `    ``// Update res` `    ``res = ((modInv % M) * (F_N % M)) % M;`   `    ``return` `res;` `}`   `// Driver code` `int` `main()` `{`   `    ``long` `long` `N = 3;` `    ``long` `long` `F_N = 6;` `    ``cout << F_1(N, F_N);` `}`

## Java

 `// Java program to implement` `// the above approach` `import` `java.util.*;` `class` `GFG{`   `static` `final` `int` `M = ``1000000007``;`   `// Function to find the ` `// value of power(X, N) % M` `static` `long` `power(``long` `x,` `                  ``long` `N)` `{` `  ``// Stores the value` `  ``// of (X ^ N) % M` `  ``long` `res = ``1``;`   `  ``// Calculate the value ` `  ``// of power(x, N) % M` `  ``while` `(N > ``0``) ` `  ``{` `    ``// If N is odd` `    ``if` `(N % ``2` `== ``1``) ` `    ``{` `      ``// Update res` `      ``res = (res * x) % M;` `    ``}`   `    ``// Update x` `    ``x = (x * x) % M;`   `    ``// Update N` `    ``N = N >> ``1``;` `  ``}` `  ``return` `res;` `}`   `// Function to find modulo ` `// multiplicative inverse ` `// of X under modulo M` `static` `long` `moduloInverse(``long` `X)` `{` `  ``return` `power(X, M - ``2``);` `}`   `// Function to find the ` `// value of F(1)` `static` `long` `F_1(``long` `N, ` `                ``long` `F_N)` `{` `  ``// Stores power(2, N - 1)` `  ``long` `P_2 = power(``2``, N - ``1``);`   `  ``// Stores modulo multiplicative` `  ``// inverse of P_2 under modulo M` `  ``long` `modInv = moduloInverse(P_2);`   `  ``// Stores the value of F(1)` `  ``long` `res;`   `  ``// Update res` `  ``res = ((modInv % M) * ` `         ``(F_N % M)) % M;`   `  ``return` `res;` `}`   `// Driver code` `public` `static` `void` `main(String[] args)` `{` `  ``long` `N = ``3``;` `  ``long` `F_N = ``6``;` `  ``System.out.print(F_1(N, F_N));` `}` `}`   `// This code is contributed by Rajput-Ji`

## Python3

 `# Python3 program to implement` `# the above approach` `M ``=` `1000000007`   `# Function to find the value` `# of power(X, N) % M` `def` `power(x, N):` `    `  `    ``# Stores the value` `    ``# of (X ^ N) % M` `    ``res ``=` `1`   `    ``# Calculate the value of` `    ``# power(x, N) % M` `    ``while` `(N > ``0``):`   `        ``# If N is odd` `        ``if` `(N & ``1``):`   `            ``# Update res` `            ``res ``=` `(res ``*` `x) ``%` `M`   `        ``# Update x` `        ``x ``=` `(x ``*` `x) ``%` `M`   `        ``# Update N` `        ``N ``=` `N >> ``1` `        `  `    ``return` `res`   `# Function to find modulo multiplicative` `# inverse of X under modulo M` `def` `moduloInverse(X):` `    `  `    ``return` `power(X, M ``-` `2``)`   `#Function to find the value of F(1)` `def` `F_1(N, F_N):`   `    ``# Stores power(2, N - 1)` `    ``P_2 ``=` `power(``2``, N ``-` `1``)`   `    ``# Stores modulo multiplicative` `    ``# inverse of P_2 under modulo M` `    ``modInv ``=` `moduloInverse(P_2)`   `    ``# Stores the value of F(1)` `    ``res ``=` `0`   `    ``# Update res` `    ``res ``=` `((modInv ``%` `M) ``*` `(F_N ``%` `M)) ``%` `M`   `    ``return` `res`   `# Driver code` `if` `__name__ ``=``=` `'__main__'``:` `    `  `    ``N ``=` `3` `    ``F_N ``=` `6` `    `  `    ``print``(F_1(N, F_N))`   `# This code is contributed by mohit kumar 29`

## C#

 `// C# program to implement` `// the above approach` `using` `System;` `class` `GFG{`   `static` `readonly` `int` `M = 1000000007;`   `// Function to find the ` `// value of power(X, N) % M` `static` `long` `power(``long` `x,` `                  ``long` `N)` `{` `  ``// Stores the value` `  ``// of (X ^ N) % M` `  ``long` `res = 1;`   `  ``// Calculate the value ` `  ``// of power(x, N) % M` `  ``while` `(N > 0) ` `  ``{` `    ``// If N is odd` `    ``if` `(N % 2 == 1) ` `    ``{` `      ``// Update res` `      ``res = (res * x) % M;` `    ``}`   `    ``// Update x` `    ``x = (x * x) % M;`   `    ``// Update N` `    ``N = N >> 1;` `  ``}` `  ``return` `res;` `}`   `// Function to find modulo ` `// multiplicative inverse ` `// of X under modulo M` `static` `long` `moduloInverse(``long` `X)` `{` `  ``return` `power(X, M - 2);` `}`   `// Function to find the ` `// value of F(1)` `static` `long` `F_1(``long` `N, ` `                ``long` `F_N)` `{` `  ``// Stores power(2, N - 1)` `  ``long` `P_2 = power(2, N - 1);`   `  ``// Stores modulo multiplicative` `  ``// inverse of P_2 under modulo M` `  ``long` `modInv = moduloInverse(P_2);`   `  ``// Stores the value of F(1)` `  ``long` `res;`   `  ``// Update res` `  ``res = ((modInv % M) * ` `         ``(F_N % M)) % M;`   `  ``return` `res;` `}`   `// Driver code` `public` `static` `void` `Main(String[] args)` `{` `  ``long` `N = 3;` `  ``long` `F_N = 6;` `  ``Console.Write(F_1(N, F_N));` `}` `}`   `// This code is contributed by shikhasingrajput`

Output:

```500000005

```

Time Complexity: O(log2N)
Auxiliary Space: O(1)

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