Find the sum of n terms of the series 12, 105, 1008, 10011, ...

Given a positive integer n. Find the sum of the first n term of the series 

12, 105, 1008, 10011, .....

Examples:

Input: n = 4
Output: 11136

Input: n = 7
Output: 11111187

Approach:

The sequence is formed by using the following pattern. For any value N-

S_{n} = \frac{10}{9}*(10^{n}-1) + \frac{n}{2}[3n+1]

The above solution can be derived following a series of steps:

Given Series-

12 + 105 + 1008 + 10011 +.......

10 + 2 + 100 + 5 + 1000 + 8 + 10000 + 11 +........

(10 + 100 + 1000 + 10000+......) + (2 + 5 + 8 + 11+......)     -(1)

The first term in the above equation is Geometric progression and the second term is Arithmetic progression.

G.P. = a*\frac{r^{n}-1}{r-1}               
where a is the first term a, r is the common ratio and n is the number of terms.

A.P. = \frac{n}{2}[2a +(n - 1)d]               
where a is the first term a, a is the common difference and n is the number of terms.

So after substituting values in equation of G.P. and A.P. and substituting corresponding equations in equation (1) we get,

10*\frac{10^{n}-1}{10-1} + \frac{n}{2}[2*2+(n-1)*3]

\frac{10}{9}*10^{n}-1 + \frac{n}{2}[4+3n-3]

\frac{10}{9}*(10^{n}-1) + \frac{n}{2}[3n+1]

So,  S_{n} = \frac{10}{9}*(10^{n}-1) + \frac{n}{2}[3n+1]

Illustration:

Input: n = 4
Output: 11136
Explanation:
S_{n} = \frac{10}{9}*(10^{4}-1) + \frac{4}{2}[3*4+1]                
S_{n} = \frac{10}{9}*(9999) + \frac{4}{2}[13]               
S_{n} = 11110 + 26               
S_{n} = 11136

This gives ans 11136.

Below is the implementation of the above approach:

// C++ program to implement
// the above approach
#include <bits/stdc++.h>
#define ll long long
using namespace std;

// Function to return sum of
// N term of the series

ll findSum(ll n)
{
    ll x = 10 * (pow(10, n) - 1) / 9;
    ll y = n * (3 * n + 1) / 2;

    return x + y;
}

// Driver Code

int main()
{
    ll n = 4;
    cout << findSum(n);
    return 0;
}

// Java program to implement
// the above approach
class GFG 
{

  // Function to return sum of
  // N term of the series
  static int findSum(int n) {
    int x = (int)(10 * (Math.pow(10, n) - 1) / 9);
    int y = n * (3 * n + 1) / 2;

    return x + y;
  }

  // Driver Code
  public static void main(String args[]) {
    int n = 4;
    System.out.println(findSum(n));
  }
}

// This code is contributed by saurabh_jaiswal.

# Python program to implement
# the above approach
# include <bits/stdc++.h>
# define ll long long

# Function to return sum of
# N term of the series
def findSum(n):
    x = 10 * ((10 ** n) - 1) / 9
    y = n * (3 * n + 1) / 2

    return int(x + y)

# Driver Code
n = 4
print(findSum(n))

# This code is contributed by saurabh_jaiswal.

// C# program to implement
// the above approach
using System;
class GFG
{

  // Function to return sum of
  // N term of the series
  static int findSum(int n) {
    int x = (int)(10 * (Math.Pow(10, n) - 1) / 9);
    int y = n * (3 * n + 1) / 2;

    return x + y;
  }

  // Driver Code
  public static void Main()
  {
    int n = 4;
    Console.Write(findSum(n));

  }
}

// This code is contributed by Samim Hossain Mondal.

  <script>
        // JavaScript code for the above approach

        // Function to return sum of
        // N term of the series
        function findSum(n) {

            let x = 10 * (Math.pow(10, n) - 1) / 9;
            let y = n * (3 * n + 1) / 2;

            return Math.floor(x) + Math.floor(y);
        }
        // Driver Code

        // Get the value of N
        let N = 4;
        document.write(findSum(N));

  // This code is contributed by Potta Lokesh
    </script>

11136

Time Complexity: O(logN) since it is using pow function

Auxiliary Space: O(1), since no extra space has been taken.