Question

Write a program to find and print the sum of the first n cubes in the following four ways.

Answer 1

Answer:

The solution code is written in Java.

1. public class Main {
3.    public static void main(String[] args) {
4.           System.out.print("Please enter a number: ");
5.           Scanner console = new Scanner(System.in);
6.           int n = console.nextInt();
7.           calculateSum1(n);
8.           calculateSum2(n);
9.           calculateSum3(n);
10.           calculateSum4(n);
11.    }
13.    public static void calculateSum1 (int num){
14.        int sum1 = 0;
16.        for(int i=1; i <= num; i++){
17.            sum1 += i*i*i;
18.        }
20.        System.out.println("Sum 1 : " + sum1);
21.    }
23.    public static void calculateSum2 (int num){
24.        int sum2 = num * num * (num + 1) * (num + 1) / 4;
25.        System.out.println("Sum 2 : " + sum2);
26.    }
28.    public static void calculateSum3 (int num){
29.        int sum3 = 0;
31.        for(int i=1; i <=num; i++){
32.            sum3 += i;
33.        }
35.        sum3 = sum3 * sum3;
37.        System.out.println("Sum 3 : " + sum3);
38.    }
40.    public static void calculateSum4 (int num){
41.        int sum4 = (num * (num + 1) * (2*num + 1)) / (4+2);
42.        System.out.println("Sum 4 : " + sum4);
43.    }
45. }

Explanation:

Firstly we create four different methods that calculate the sum of the first n cubes in different ways (Line 13 - 43).

This is important to understand the concept of  operator precedence to work out the different ways of summation.

For example, The operator * and / will always have higher precedence than +.

On another hand, any expression enclosed within the parenthesis will have highest precedence and therefore the value will be evaluated prior to others.  

This is also important to understand the expression evaluation will always start from left to right.