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: 11136Input: 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++ 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>
|
Output
11136
Time Complexity: O(logN) since it is using pow function
Auxiliary Space: O(1), since no extra space has been taken.