# Find the sum of infinite series 1^2.x^0 + 2^2.x^1 + 3^2.x^2 + 4^2.x^3 +…….

Given an infinite series and a value x, the task is to find its sum. Below is the infinite series

1^2*x^0 + 2^2*x^1 + 3^2*x^2 + 4^2*x^3 +……. upto infinity, where x belongs to (-1, 1)

Examples:

Input: x = 0.5
Output: 12

Input: x = 0.9
Output: 1900


## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:

Though the given series is not an Arithmetico-Geometric series, however, the differences and so on, forms an AP. So, we can use the Method of Differences. Hence, the sum will be (1+x)/(1-x)^3.

Below is the implementation of above approach:

## C++

 // C++ implementation of above approach  #include  #include     using namespace std;     // Function to calculate sum  double solve_sum(double x)  {      // Return sum      return (1 + x) / pow(1 - x, 3);  }     // Driver code  int main()  {      // declaration of value of x      double x = 0.5;         // Function call to calculate      // the sum when x=0.5      cout << solve_sum(x);         return 0;  }

## Java

 // Java Program to find   //sum of the given infinite series  import java.util.*;     class solution  {  static double calculateSum(double x)  {         // Returning the final sum  return (1 + x) / Math.pow(1 - x, 3);     }     //Driver code  public static void main(String ar[])  {           double x=0.5;    System.out.println((int)calculateSum(x));     }  }  //This code is contributed by Surendra_Gangwar

## Python

 # Python implementation of above approach     # Function to calculate sum  def solve_sum(x):      # Return sum      return (1 + x)/pow(1-x, 3)     # driver code     # declaration of value of x  x = 0.5    # Function call to calculate the sum when x = 0.5  print(int(solve_sum(x)))

## C#

 // C# Program to find sum of  // the given infinite series  using System;     class GFG  {  static double calculateSum(double x)  {         // Returning the final sum  return (1 + x) / Math.Pow(1 - x, 3);     }     // Driver code  public static void Main()  {      double x = 0.5;      Console.WriteLine((int)calculateSum(x));  }  }     // This code is contributed  // by inder_verma..

## PHP

 

Output:

12


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