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I was trying to solve one of the question. Could anyone please let me know how I can make this code much effective in terms of time and space complexity?

You are just learning to code and are finished with loops and
functions. Now, you are given the following pseudocode:

/*
 * The function has two integer parameters: k and n
 * The function returns the value of sum
 */
 function f(k, n) 
 sum = 0;

 for (i = 1; i ≤ n; i += 1) 
 sum += i;
 i *= k;
 

 return sum;

For three given integers k,a, and b, find the value of S mod (10^9 +
7), where S is defined as:

$$s=sum_n=a^b▒〖f(k,n)$$

Input Format

The first line of the input is an integer Q, the total number of
queries. Each of the next Q lines contains three space separated
integers k,a,and b

Output Format

For each query, print the value of S mod (10^9 + 7) on a new line.

Sample Input

4
2 1 5
3 1 5
4 1 5
5 1 5

Sample Output

14
13
10
5

Explanation

Query 2 1 5
f(2,1) = 1
f(2,2) = 1
f(2,3) = 4
f(2,4) = 4
f(2,5) = 4

So, s = f(2,1)+f(2,2)+f(2,3)+f(2,4)+f(2,5) = 14

 Query 3 1 5

 f(3,1) = 1
 f(3,2) = 1
 f(3,3) = 1
 f(3,4) = 5
 f(3,5) = 5

So, s = f(3,1)+f(3,2)+f(3,3)+f(3,4)+f(3,5) = 13

 Query 4 1 5
 f(4,1) = 1
 f(4,2) = 1
 f(4,3) = 1
 f(4,4) = 1
 f(4,5) = 6

So, s = f(4,1)+f(4,2)+f(4,3)+f(4,4)+f(4,5) = 10

 Query 5 1 5
 f(5,1) = 1
 f(5,2) = 1
 f(5,3) = 1
 f(5,4) = 1
 f(5,5) = 1

So, s = f(5,1)+f(5,2)+f(5,3)+f(5,4)+f(5,5) = 5

import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class maxSum 

static int simplestSum(int k, int a, int b) 
 int count = 1;
 int totalSum = 0;
 while(count <= b) 
 int z = maxSum(k, a);
 int mul = k*a;
 if(mul < b) 
 totalSum = totalSum + (z*(k*a));
 else 
 totalSum = totalSum + (z*((b-count)+1));
 

 count = count + (k*a);
 a = count;
 
 return totalSum;


static int maxSum(int k , int a) 
 int sum = 0;
 for (int i = 1; i <= a; i += 1) 
 sum += i;
 i *= k;
 
 return sum;


public static void main(String args) 
 Scanner in = new Scanner(System.in);
 int q = in.nextInt();
 for(int a0 = 0; a0 < q; a0++)
 int k = in.nextInt();
 int a = in.nextInt();
 int b = in.nextInt();
 int result = simplestSum(k, a, b);
 System.out.println(result);
 
 in.close();
 

 function f(k, n) 
 summ = 0
 max = n*k;
 for (i = 0; i < max; i += k)
 summ += i

 return sum;

or even simpler, using the formula of summ:

summ = k + 2k + ... + nk = (k + nk) + (2k + (n-1)k) +... = k*(n+1)*n/2

function f(k, n) 
return (n+1)*n/2 *k;