Skip to content
Related Articles

Related Articles

Improve Article

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 : 22 Mar, 2021
Geek Week

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 (2^2-1^2), (3^2-2^2), ...  and so on, forms an AP. So, we can use the Method of Differences.
Let\: S = 1 + 4x + 9x^2 + 16x^3 + ...\infty \\\\ Multiply\: both\: sides\: with\: common\: ratio\: x\: of\: the\: GP(geometric progression).\\ S_x = x + 4x^2 + 9x^3 + ...\infty \\ \\ Now, \: subtract\: the\: two\: equations.\\ => (1-x)S = 1 + 3x + 5x^2 + 7x^3 + ...\infty \null\hfill (1)\\\\ Now, \: let\: R = 1 + 3x + 5x^2 + 7x^3 + ...\infty, \: which\: is\: an \:Arithmetico-Geometric\: series\: with \:a=1, \: d=2 \:and \:r=x.\\ For an A.G.P., $Sum \:R \:= \frac{a}{1-r} + \frac{rd}{(1-r)^2} \\ Substituting\: the \:values, \:we\: get\: R = \frac{1+x}{(1-x)^2} \\ Substitute\: R\: in\: (1), \:we\: get, (1-x)S=\frac{1+x}{(1-x)^2} \\ => S= \frac{1+x}{(1-x)^3}$
Hence, the sum will be (1+x)/(1-x)^3.
Below is the implementation of above approach: 
 



C++




// C++ implementation of above approach
#include <iostream>
#include <math.h>
 
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




<?php
// PHP implementation of
// above approach
 
// Function to calculate sum
function 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
echo solve_sum($x);
 
// This code is contributed
// by inder_verma
?>

Javascript




<script>
// javascript Program to find
//sum of the given infinite series
 
 
function calculateSum(x)
{
     
// Returning the final sum
return (1 + x) / Math.pow(1 - x, 3);
 
}
 
//Driver code
  
var x=0.5;
document.write(parseInt(calculateSum(x)));
 
// This code is contributed by 29AjayKumar
 
</script>
Output: 
12

 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.




My Personal Notes arrow_drop_up
Recommended Articles
Page :