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Find the sum of n terms of the series 1 , 2a , 3a2 , 4a3 , 5a4 , …

  • Last Updated : 23 Dec, 2021

Given a series .1, 2a, 3a^{2}, 4a^{3}, 5a^{4},.....                        and the value of a. Find the sum of the first n term of the series.

Examples:

Input: a = 3, n = 4
Output: 142

Input: a = 5, n = 1
Output: 1

Brute Force Approach:

A simple approach can be iterating N terms of the series and adding them to calculate the sum for any value of a. Follow the steps below to understand the approach:

For each iteration:

  1. Calculate an [ n = 0 ].
  2. Multiply an with (n+1).
  3. Add (n+1)*an to sum and increment n by 1.
  4. Repeat the above processes n times.

Illustration:

a = 3 and n = 4

Loop will be executed n number of times i.e 4 in this case.

Loop 1: Initially the value of a = 1, n = 0, sum = 0

  1. an = 30
          = 1
  2. an * (n+1) = 30 * (0+1)
                    = 1 * (1)
                    = 1
  3. sum = sum + an * (n+1)
            = 0 + 1
            = 1
  4. Increment n by 1.

Loop 2: The value of a = 3, n = 1, sum = 1

  1. an = 31
         = 3
  2. an * (n+1) = 31 * (1+1)
                    = 3 * (2)
                    = 6
  3. sum = sum + an * (n+1)
           = 1 + 6
           = 7
  4. Increment value of n by 1.

Loop 3: The value of a = 3, n = 2, sum = 7

  1. an = 32
         = 9
  2. an * (n+1) = 32 * (2+1)
                     = 9 * (3)
                     = 27
  3. sum = sum + an * (n+1)
            = 7 + 27
            = 34
  4. Increment n by 1.

Loop 4: The value of a = 3, n = 3, sum = 34

  1. an = 33
          = 27
  2. an * (n+1) = 33 * (3+1)
                     = 27 * (4)
                     = 108
  3. sum = sum + an * (n+1)
            = 34 + 108
            = 142
  4. Increment the value of n by 1.

Below is the implementation of the above approach: 

C++




// C++ implementation for the
// approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to calculate
// the sum
void calcSum(int a, int n)
{
  // Edge Cases
  if (n < 0)
  {
    cout << "Invalid Input";
    return;
  }
 
  if (a == 0 || n == 1)
  {
    cout << 1;
    return;
  }
 
  // Initialize the variables
  int Sum = 0;
 
  // Calculate Sum upto N terms
  for(int i = 0; i < n; i++)
  {
    int r = pow(a, (i)) * (i + 1);
    Sum += r;
  }
 
  // Print Sum
  cout << Sum;
}
 
// Driver Code
int main()
{
  int a = 3;
  int n = 4;
   
  // Invoke calcSum function with
  // values of a and n
  calcSum(a, n);
  return 0;
}

Java




// Java implementation for the
// approach
import java.util.*;
 
class GFG{
 
// Function to calculate
// the sum
static void calcSum(int a, int n)
{
   
  // Edge Cases
  if (n < 0)
  {
    System.out.print("Invalid Input");
    return;
  }
 
  if (a == 0 || n == 1)
  {
    System.out.print(1);
    return;
  }
 
  // Initialize the variables
  int Sum = 0;
 
  // Calculate Sum upto N terms
  for(int i = 0; i < n; i++)
  {
    int r = (int) (Math.pow(a, (i)) * (i + 1));
    Sum += r;
  }
 
  // Print Sum
  System.out.print(Sum);
}
 
// Driver Code
public static void main(String[] args)
{
  int a = 3;
  int n = 4;
   
  // Invoke calcSum function with
  // values of a and n
  calcSum(a, n);
}
}
 
// This code is contributed by 29AjayKumar

Python3




# Python 3 implementation for the
# approach
 
# Function to calculate
# the sum
def calcSum(a, n):
 
    # Edge Cases
    if (n < 0):
        print("Invalid Input")
        return
 
    if (a == 0 or n == 1):
 
        print(1)
        return
 
    # Initialize the variables
    Sum = 0
 
    # Calculate Sum upto N terms
    for i in range(n):
        r = pow(a, (i)) * (i + 1)
        Sum += r
 
    # Print Sum
    print(Sum)
 
# Driver Code
if __name__ == "__main__":
 
    a = 3
    n = 4
 
    # Invoke calcSum function with
    # values of a and n
    calcSum(a, n)
 
    # This code is contributed by ukasp.

C#




// C# program to find GCD of two
// numbers
using System;
using System.Collections;
 
class GFG {
 
// Function to calculate
// the sum
static void calcSum(int a, int n)
{
  // Edge Cases
  if (n < 0)
  {
    Console.Write("Invalid Input");
    return;
  }
 
  if (a == 0 || n == 1)
  {
    Console.Write(1);
    return;
  }
 
  // Initialize the variables
  int Sum = 0;
 
  // Calculate Sum upto N terms
  for(int i = 0; i < n; i++)
  {
    int r = (int)Math.Pow(a, (i)) * (i + 1);
    Sum += r;
  }
 
  // Print Sum
  Console.Write(Sum);
}
 
// Driver method
public static void Main()
{
    int a = 3;
    int n = 4;
   
    // Invoke calcSum function with
    // values of a and n
    calcSum(a, n);
}
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript




<script>
 
// JavaScript implementation for the 
// approach
 
// Function to calculate
// the sum
function calcSum(a, n)
{
     
    // Edge Cases
    if (n < 0)
    {
        document.write("Invalid Input");
        return;
    }
 
    if (a == 0 || n == 1)
    {
        document.write(1);
        return;
    }
 
    // Initialize the variables
    let Sum = 0;
 
    // Calculate Sum upto N terms
    for(let i = 0; i < n; i++)
    {
        let r = Math.pow(a, (i)) * (i + 1);
        Sum += r;
    }
 
    // Print Sum
    document.write(Sum);
}
 
// Driver Code
let a = 3;
let n = 4;
 
// Invoke calcSum function with
// values of a and n
calcSum(a, n);
 
// This code is contributed by Potta Lokesh
 
</script>

Output:

142

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

Efficient Approach

In this approach, an efficient solution is proposed using the concept of Geometric progression. The sum of the series of n terms in a Geometric Progression (G.P.) with first term a and common ratio r is given as:

S_{n} = \frac{a(r^{n}-1)}{r - 1}

Let’s use this concept to reach a solution to the problem.

Let S = 1 +2a + 3a^{2} + 4a^{3} +............+n^{th} term

Clearly nth term is na^{n-1}

S = 1 +2a + 3a^{2} + 4a^{3} +............+na^{n-1}.............                       .     (1)

Multiply both sides with ‘a’, we get,

Sa = (1 +2a + 3a^{2} + 4a^{3} +............+na^{n-1})a

Sa = 0 + a +2a^{2} + 3a^{3} + 4a^{4} +............+ (n - 1)a^{n - 1} + na^{n}............                           (2)

Subtracting equation (2) from (1), we get

S - Sa = (1 +2a + 3a^{2} + 4a^{3} +............+na^{n-1}) - (0 + a +2a^{2} + 3a^{3} + 4a^{4} +............+ na^{n})

S(1 - a) = 1 + a + a^{2} + a^{3} +..............+ a^{n - 1} - na^{n}

Clearly this is the Geometric Progression (G.P.) of n terms with first term 1 and common ration a.

G.P. of n terms with first term a and common ratio r is:

S_{n} = \frac{a(r^{n}-1)}{r - 1}

Using the above formula, we have

S(1 - a) = \frac{1(a^{n} - 1)}{(a - 1)} - na^{n}

Dividing both sides by (1 – a), we get

\frac{S(1 - a)}{(1 - a)} = \frac{1(a^{n} - 1)}{(a - 1)} * \frac{1}{(1 - a)} - na^{n} * \frac{1}{(1 - a)}

S = \frac{-1(1 - a^{n})}{(a - 1)} * \frac{-1}{(a - 1)} - \frac{na^{n}}{(1 - a)}

S = \frac{1(1 - a^{n})}{(a - 1)^{2}} - \frac{na^{n}}{(1 - a)}

Therefore, the sum of the series S = 1 +2a + 3a^{2} + 4a^{3} +............+n^{th} term                        is 

\frac{(1 - a^{n})}{(a - 1)^{2}} - \frac{na^{n}}{(1 - a)}

For a != 1 the formula for sum of the series is:

S = \frac{(1 - a^{n})}{(a - 1)^{2}} - \frac{na^{n}}{(1 - a)}

For a = 1 the formula for sum of the series is:
The series reduces to sum of first n natural numbers and the formula becomes-

S = \frac{n(n + 1)}{2}

Illustration:

For a  = 3, n = 4

Since a != 1, therefore use the formula

S = \frac{(1 - a^{n})}{(a - 1)^{2}} - \frac{na^{n}}{(1 - a)}

Substituting the values of a and n in the above formula, we get

S = \frac{(1 - 3^{4})}{(3 - 1)^{2}} - \frac{4 * 3^{4}}{(1-3)}

S = \frac{1 - 81}{2^{2}} - \frac{324}{-2}

S = -20 – (-162)

S = 142

So, the sum of the series S = 1 +2a + 3a^{2} + 4a^{3} +............+n^{th} term                        with value of a = 3 and n = 4 is 142.

Below is the implementation of the above approach: 

C++




// C++ program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to calculate
// the sum
void calcSum(int a, int n)
{
  // Edge Cases
  if (n < 0)
  {
    cout << "Invalid Input";
    return;
  }
 
  if (a == 0 || n == 1)
  {
    cout << 1;
    return;
  }
 
  // Sum of First N Natural Numbers
  // In case a = 1
  if (a == 1)
  {
    // Avoiding Overflow
    if (n % 2 == 0)
      cout << (n / 2) * (n + 1);
 
    else
      cout << ((n + 1) / 2) * n;
  }
 
  // Calculate Sum with the help
  // of formula
  int r = pow(a, n);
  int d = pow(a - 1, 2);
  int Sum = (1 - r * (1 + n - n * a)) / d;
 
  // Print Sum
  cout << Sum;
}
 
// Driver Code
int main()
{
  int a = 3;
  int n = 4;
   
  // Invoke calcSum function
  // with values of a and n
  calcSum(a, n);
  return 0;
}

Java




// Java program to implement
// the above approach
class GFG {
 
    // Function to calculate
    // the sum
    public static void calcSum(int a, int n)
    {
       
        // Edge Cases
        if (n < 0) {
            System.out.println("Invalid Input");
            return;
        }
 
        if (a == 0 || n == 1) {
            System.out.println(1);
            return;
        }
 
        // Sum of First N Natural Numbers
        // In case a = 1
        if (a == 1) {
            // Avoiding Overflow
            if (n % 2 == 0)
                System.out.println((n / 2) * (n + 1));
 
            else
                System.out.println(((n + 1) / 2) * n);
        }
 
        // Calculate Sum with the help
        // of formula
        int r = (int) Math.pow(a, n);
        int d = (int) Math.pow(a - 1, 2);
        int Sum = (1 - r * (1 + n - n * a)) / d;
 
        // Print Sum
        System.out.println(Sum);
    }
 
    // Driver Code
    public static void main(String args[]) {
        int a = 3;
        int n = 4;
 
        // Invoke calcSum function
        // with values of a and n
        calcSum(a, n);
    }
}
 
// This code is contributed by saurabh_jaiswal.

Python3




# Python program to implement
# the above approach
 
# Function to calculate
# the sum
def calcSum(a, n):
   
    # Edge Cases
    if (n < 0):
        print("Invalid Input");
        return;
 
    if (a == 0 or n == 1):
        print(1);
        return;
 
    # Sum of First N Natural Numbers
    # In case a = 1
    if (a == 1):
       
        # Avoiding Overflow
        if (n % 2 == 0):
            print((n // 2) * (n + 1));
 
        else:
            print(((n + 1) // 2) * n);
 
    # Calculate Sum with the help
    # of formula
    r =  pow(a, n);
    d = pow(a - 1, 2);
    Sum = (1 - r * (1 + n - n * a)) // d;
 
    # PrSum
    print(Sum);
 
# Driver Code
if __name__ == '__main__':
    a = 3;
    n = 4;
 
    # Invoke calcSum function
    # with values of a and n
    calcSum(a, n);
 
# This code is contributed by 29AjayKumar

C#




// C# program to implement
// the above approach
 
using System;
class GFG {
 
    // Function to calculate
    // the sum
    public static void calcSum(int a, int n)
    {
       
        // Edge Cases
        if (n < 0) {
            Console.WriteLine("Invalid Input");
            return;
        }
 
        if (a == 0 || n == 1) {
            Console.WriteLine(1);
            return;
        }
 
        // Sum of First N Natural Numbers
        // In case a = 1
        if (a == 1) {
            // Avoiding Overflow
            if (n % 2 == 0)
                Console.WriteLine((n / 2) * (n + 1));
 
            else
                Console.WriteLine(((n + 1) / 2) * n);
        }
 
        // Calculate Sum with the help
        // of formula
        int r = (int) Math.Pow(a, n);
        int d = (int) Math.Pow(a - 1, 2);
        int Sum = (1 - r * (1 + n - n * a)) / d;
 
        // Print Sum
        Console.WriteLine(Sum);
    }
 
    // Driver Code
    public static void Main() {
        int a = 3;
        int n = 4;
 
        // Invoke calcSum function
        // with values of a and n
        calcSum(a, n);
    }
}
 
// This code is contributed by gfgking.

Javascript




<script>
// Javascript program to implement
// the above approach
 
// Function to calculate
// the sum
function calcSum(a, n)
{
  // Edge Cases
  if (n < 0)
  {
    document.write("Invalid Input");
    return;
  }
 
  if (a == 0 || n == 1)
  {
    document.write(1);
    return;
  }
 
  // Sum of First N Natural Numbers
  // In case a = 1
  if (a == 1)
  {
    // Avoiding Overflow
    if (n % 2 == 0)
      document.write((n / 2) * (n + 1));
 
    else
      document.write(((n + 1) / 2) * n);
  }
 
  // Calculate Sum with the help
  // of formula
  let r = Math.pow(a, n);
  let d = Math.pow(a - 1, 2);
  let Sum = (1 - r * (1 + n - n * a)) / d;
 
  // Print Sum
  document.write(Sum);
}
 
// Driver Code
let a = 3;
let n = 4;
   
// Invoke calcSum function
// with values of a and n
calcSum(a, n);
 
// This code is contributed by Samim Hossain Mondal.
</script>

Output:

142

Time Complexity: O(1)

Auxiliary Space: O(1)


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