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Program to find the sum of the series 1 + x + x^2+ x^3+ .. + x^n
  • Last Updated : 12 Jun, 2020

Given an integer X, the task is to print the series and find the sum of the series 1 + x^{1} + x^{2} + x^{3} + x^{4} ... x^{N}

Examples :

Input: X = 2, N = 5
Output: Sum = 31
1 2 4 8 16

Input: X = 1, N = 10
Output: Sum = 10
1 1 1 1 1 1 1 1 1 1

Approach: The idea is to traverse over the series and compute the sum of the N terms of the series. The Nth term of the series can be computed as:



Nth Term = (N-1)th Term * X

Below is the implementation of the above approach:

C++

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// C++ implementatio to find
// sum of series of
// 1 + x^2 + x^3 + ....+ x^n
  
#include <bits/stdc++.h>
  
using namespace std;
  
// Function to find the sum of
// the series and print N terms
// of the given series
double sum(int x, int n)
{
    double i, total = 1.0, multi = x;
  
    // First Term
    cout << total << " ";
  
    // Loop to print the N terms
    // of the series and find their sum
    for (i = 1; i < n; i++) {
  
        total = total + multi;
        cout << multi << " ";
        multi = multi * x;
    }
  
    cout << "\n";
    return total;
}
  
// Driver code
int main()
{
    int x = 2;
    int n = 5;
    cout << fixed
         << setprecision(2)
         << sum(x, n);
    return 0;
}

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C

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// C implemenation to find the sum
// of series 1 + x^2 + x^3 + ....+ x^n
  
#include <math.h>
#include <stdio.h>
  
// Function to print the sum
// of the series
double sum(int x, int n)
{
    double i, total = 1.0, multi = x;
  
    // First Term of series
    printf("1 ");
  
    // Loop to find the N
    // terms of the series
    for (i = 1; i < n; i++) {
  
        total = total + multi;
        printf("%.1f ", multi);
        multi = multi * x;
    }
    printf("\n");
    return total;
}
  
// Driver Code
int main()
{
    int x = 2;
    int n = 5;
    printf("%.2f", sum(x, n));
    return 0;
}

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Java

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// Java implementation to find
// the sum of series
// 1 + x^2 + x^3 + ....+ x^n
  
class GFG {
  
    // Java code to print the sum
    // of the given series
    static double sum(int x, int n)
    {
        double i, total = 1.0, multi = x;
  
        // First Term
        System.out.print("1 ");
  
        // Loop to print the N terms
        // of the series and compute
        // their sum
        for (i = 1; i < n; i++) {
            total = total + multi;
            System.out.print(multi);
            System.out.print(" ");
            multi = multi * x;
        }
  
        System.out.println();
        return total;
    }
  
    // Driver Code
    public static void main(String[] args)
    {
        int x = 2;
        int n = 5;
        System.out.printf(
            "%.2f", sum(x, n));
    }
}

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Python3

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# Python3 program to find sum of 
# series of 1 + x^2 + x^3 + ....+ x^n 
  
# Function to find the sum of 
# the series and print N terms 
# of the given series 
def sum(x, n): 
      
    total = 1.0
    multi =
  
    # First Term 
    print(1, end = " "
  
    # Loop to print the N terms 
    # of the series and find their sum 
    for i in range(1, n): 
          
        total = total + multi 
        print('%.1f' % multi, end = " "
        multi = multi *
          
    print('\n')
    return total; 
  
# Driver code 
x = 2
n = 5
print('%.2f' % sum(x, n))
  
# This code is contributed by Pratik Basu 

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C#

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// C# implementation to find
// the sum of series
// 1 + x^2 + x^3 + ....+ x^n
using System;
class GFG{
  
// C# code to print the sum
// of the given series
static double sum(int x, int n)
{
    double i, total = 1.0, multi = x;
  
    // First Term
    Console.Write("1 ");
  
    // Loop to print the N terms
    // of the series and compute
    // their sum
    for (i = 1; i < n; i++)
    {
        total = total + multi;
        Console.Write(multi);
        Console.Write(" ");
        multi = multi * x;
    }
    Console.WriteLine();
    return total;
}
  
// Driver Code
public static void Main(String[] args)
{
    int x = 2;
    int n = 5;
    Console.Write("{0:F2}", sum(x, n));
}
}
  
// This code is contributed by Rajput-Ji

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Output:

1 2.0 4.0 8.0 16.0 
31.00

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