# Find sum of the series 1+22+333+4444+…… upto n terms

• Difficulty Level : Medium
• Last Updated : 25 May, 2022

Given a number N. The task is to find the sum of the below series up to N-th term:

1 + 22 + 333 + 4444 + …up to n terms

Examples

Input: N = 3
Output: 356

Input: N = 10
Output: 12208504795

Approach:

Below is the implementation of the above approach:

## C++

 // CPP program to find the sum// of given series #include #include  using namespace std; // Function to calculate sumint findSum(int n){    // Return sum    return (pow(10, n + 1) * (9 * n - 1) + 10) /                    pow(9, 3) - n * (n + 1) / 18;} // Driver codeint main(){    int n = 3;     cout << findSum(n);         return 0;}

## Java

 // Java Program to find// Sum of first n termsimport java.util.*; class solution{static int calculateSum(int n){ // Returning the final sumreturn ((int)Math.pow(10, n + 1) * (9 * n - 1) + 10) /                (int)Math.pow(9, 3) - n * (n + 1) / 18;} // Driver codepublic static void main(String ar[]){// no. of terms to find the sumint n=3;System.out.println("Sum= "+ calculateSum(n)); }} //This code is contributed by Surendra_Gangwar

## Python 3

 # Python program to find the sum of given series.  # Function to calculate sumdef solve_sum(n):    # Return sum    return (pow(10, n + 1)*(9 * n - 1)+10)/pow(9, 3)-n*(n + 1)/18 # driver coden = 3 print(int(solve_sum(n)))

## C#

 // C# Program to find// Sum of first n termsusing System;class solution{static int calculateSum(int n){ // Returning the final sumreturn ((int)Math.Pow(10, n + 1) * (9 * n - 1) + 10) /                (int)Math.Pow(9, 3) - n * (n + 1) / 18;} // Driver codepublic static void  Main(){// no. of terms to find the sumint n=3;Console.WriteLine("Sum= "+ calculateSum(n)); }} //This code is contributed by inder_verma.

## PHP

 

## Javascript

 

Output:

356

Time Complexity: O(logn), where n represents the given integer, as we have used pow function.
Auxiliary Space: O(1), no extra space is required, so it is a constant.

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