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

• Last Updated : 16 Aug, 2022

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-

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.
where a is the first term a, r is the common ratio and n is the number of terms.

A.P.
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,

So,

Illustration:

Input: n = 4
Output: 11136
Explanation:

This gives ans 11136.

Below is the implementation of the above approach:

## C++

 // C++ program to implement // the above approach #include  #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

 // 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.

## Python3

 # Python program to implement # the above approach # include  # 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#

 // 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.

## Javascript

 

Output

11136

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

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

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