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

• Difficulty Level : Hard
• Last Updated : 31 Aug, 2022

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

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 sumdouble solve_sum(double x){    // Return sum    return (1 + x) / pow(1 - x, 3);} // Driver codeint 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 seriesimport java.util.*; class solution{static double calculateSum(double x){     // Returning the final sumreturn (1 + x) / Math.pow(1 - x, 3); } //Driver codepublic 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 sumdef solve_sum(x):    # Return sum    return (1 + x)/pow(1-x, 3) # driver code # declaration of value of xx = 0.5 # Function call to calculate the sum when x = 0.5print(int(solve_sum(x)))

## C#

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

## PHP

 

## Javascript

 

Output:

12

Time Complexity: O(1)

Auxiliary Space: O(1)

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