calculate the sum of all square numbers between 1 and 100 in java

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```public class SumSquares {
public static void main(String[] args){
int num = 1;
int square;
int sum = 0;
while(num <= 100){
square = (num * num);
sum = sum + square;
num++;
}
System.out.println(sum);
}
```

}

this is how i tried, in this the output is 338350, but this isn't what i want

this is what i want ==> (1+4+9+16+25+36+49+64+81)= 285

the output of the java program should be 285

Your stopping condition is wrong, since you are adding the squares from 1*1 to 100*100. You want to add squares from 1*1 to 9*9.

It should be:

```while (num < 10) {
square = (num * num);
sum = sum + square;
num++;
}
```

or

```while (num * num < 100) {
square = (num * num);
sum = sum + square;
num++;
}
```

The sum of all squares between 1 and 100 inclusive?, If you want to do it in your way I mean first square the value of each number,keep it in a variable and add it to the sum you can use a different  Sum = 5050. The above program loops from 1 to the given num(100) and adds all numbers to the variable sum.

Using `Java 9` streams you could do as follows:

```System.out.println(IntStream.range(1, Integer.MAX_VALUE)
.map(v -> v * v)
.takeWhile(v -> v < 100)
.sum());
```

Java Program for Sum of squares of first n natural numbers , Java Program for Sum of squares of first n natural numbers. Given a positive integer N. The task is to find 12 + 22 + 32 + … Java Program to find sum of 1)​2) + 3 * ( n + (n - 1) ) + 1 + 1 By putting all equation, we get (n + 1)3 = 13 + 3 * Σ k2 + 3 * Σ k + Σ 1 n3 + 3 * n2 + 3 * n + 1 = 1 overflow upto some extent for large value. Beginners Java program to find sum of odd numbers between 1 -100. Java program to find the sum of the digits of a number. Facebook. twitter. google+. pinterest.

Some alternatives :

```        int num = 1;
int square =0;
int sum = 0;
while(true){
square = (num * num);
if(square >=100) {break;}
sum = sum + square;
num++;
}
System.out.println(sum);

//using java 8 Stream
int sumUsingStream =  IntStream.range(1, 100)
.filter((v)-> (v*v) < 100 )
.map(v -> v * v)
.sum();

System.out.println(sumUsingStream);
```

Sum of squares of first n natural numbers, Naive approach : A naive approach will be to run a loop from 1 to n and sum up all the squares. Fucntion to calculate sum Java program to calculate. Write a Java Program to Calculate Sum of Even Numbers from 1 to N using For Loop, and While Loop with example. Any number that is divisible by 2 is an even number. TIP : We already explained the logic to check the number is Even or Not in Java Odd or Even Program article. I suggest you refer to the same. This Java program allows the user to

sum of squares between 1-100 ?, Hi Well i am trying to convert vb code to C# .. i know how to do it but first of all i need code for the fallowing any help please A perfect square us  Write a program in Java to find the sum of all odd numbers between 0 to N using loop. Algorithm to find sum of all odd numbers between 1 to N Take N as input from user and store it in an integer variable.

Java Program to find Sum of Natural Numbers, The positive integers 1, 2, 3, 4 etc. are known as natural numbers. Here we will see three programs to calculate and display the sum of natural numbers. As the title states, I have trouble understanding loops and have come up with a way to do a simple 1 through 100 sum, but like I said, the loops are causing me some confusion. I think I have the FO

Program to print all happy numbers between 1 to 100, To find whether a given number is happy or not, calculate the square of each digit present in number and add it to a variable sum. If resulting sum is equal to 1  Java Program to Calculate Sum of Odd Numbers Example 2. In this Java program to find the sum of odd numbers, we used for loop without the If statement. If you observe the Java code, we started i from 1 and incremented by 2 (not 1). It means for the first iteration i = 1, and for the second iteration i = 3 (not 2) so on.

• Then you should probably check if `square <= 100`.
• `while(num*num <= 100)` will fix the problem.
• Or rather `while(num <= Math.sqrt(100)) {...}`
• @CoderinoJavarino That will work too, though it should be `<` based on the question.