Given an integer **P**, that increases either by **A** or **B** with **50%** probability each, in the next **N** consecutive days, the task is to find the expected value after **N** days.

**Examples:**

Input:P = 1000, A = 5, B = 10, N = 10Output:1075Explanation:

Expected increased value after N consecutive days is equal to:

P + N * (a + b) / 2 = 1000 + 10 × 7.5 = 1000 + 75 = 1075.

Input:P = 2000, a = 10, b = 20, N = 30Output:2450

**Approach: **Follow the steps to solve the problem:

- Expected value of increase each day = (Probability of increase by A) * A + (Probability of value increase by B) * B =
**(1 / 2) * A + (1 / 2) * B**. - Therefore, increase in value after one day =
**(a + b) / 2.** - Therefore, increase in value after
**N**days =**N * (a + b) / 2.** - Therefore, increased value after
**N**days =**P + N * (a + b) / 2.** - Print the increased value as the required answer.

Below is the implementation of the above approach:

## C++

`// C++ program for` `// the above approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to find the increased` `// value of P after N days` `void` `expectedValue(` `int` `P, ` `int` `a,` ` ` `int` `b, ` `int` `N)` `{` ` ` `// Expected value of the` ` ` `// number P after N days` ` ` `double` `expValue` ` ` `= P + (N * 0.5 * (a + b));` ` ` `// Print the expected value` ` ` `cout << expValue;` `}` `// Driver Code` `int` `main()` `{` ` ` `int` `P = 3000, a = 20, b = 10, N = 30;` ` ` `expectedValue(P, a, b, N);` ` ` `return` `0;` `}` |

## Java

`// Java program for the above approach` `class` `GFG{` `// Function to find the increased` `// value of P after N days` `static` `void` `expectedValue(` `int` `P, ` `int` `a,` ` ` `int` `b, ` `int` `N)` `{` ` ` ` ` `// Expected value of the` ` ` `// number P after N days` ` ` `double` `expValue = P + (N * ` `0.5` `* (a + b));` ` ` `// Print the expected value` ` ` `System.out.print(expValue);` `}` `// Driver code` `public` `static` `void` `main(String[] args)` `{` ` ` `int` `P = ` `3000` `, a = ` `20` `, b = ` `10` `, N = ` `30` `;` ` ` `expectedValue(P, a, b, N);` `}` `}` `// This code is contributed by abhinavjain194` |

## Python3

`# Python3 program for` `# the above approach` `# Function to find the increased` `# value of P after N days` `def` `expectedValue(P, a, b, N):` ` ` ` ` `# Expected value of the` ` ` `# number P after N days` ` ` `expValue ` `=` `P ` `+` `(N ` `*` `0.5` `*` `(a ` `+` `b))` ` ` `# Print the expected value` ` ` `print` `(` `int` `(expValue))` `# Driver Code` `if` `__name__ ` `=` `=` `'__main__'` `:` ` ` ` ` `P ` `=` `3000` ` ` `a ` `=` `20` ` ` `b ` `=` `10` ` ` `N ` `=` `30` ` ` ` ` `expectedValue(P, a, b, N)` ` ` `# This code is contributed by ipg2016107` |

## C#

`// C# program for the above approach` `using` `System;` `class` `GFG{` ` ` `// Function to find the increased` `// value of P after N days` `static` `void` `expectedValue(` `int` `P, ` `int` `a,` ` ` `int` `b, ` `int` `N)` `{` ` ` ` ` `// Expected value of the` ` ` `// number P after N days` ` ` `double` `expValue = P + (N * 0.5 * (a + b));` ` ` `// Print the expected value` ` ` `Console.Write(expValue);` `}` `// Driver code` `static` `void` `Main()` `{` ` ` `int` `P = 3000, a = 20, b = 10, N = 30;` ` ` `expectedValue(P, a, b, N);` `}` `}` `// This code is contributed by abhinavjain194` |

## Javascript

`<script>` `// Javascript program for` `// the above approach` ` ` `// Function to find the increased` `// value of P after N days` ` ` `function` `expectedValue(P, a, b, N)` ` ` `{` ` ` ` ` `// Expected value of the` ` ` `// number P after N day` ` ` ` ` `var` `expValue = P + (N * 0.5 * (a + b)) ;` ` ` ` ` `return` `expValue;` ` ` `}` ` ` ` ` `// Driver code` ` ` ` ` `var` `P = 3000` ` ` `var` `a = 20` ` ` `var` `b = 10` ` ` `var` `N = 30` ` ` ` ` `document.write(expectedValue(P, a, b, N));` ` ` `</script>` |

**Output:**

3450

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

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