How to find sum of squares of first N natural numbers in Kotlin?

In this tutorial, you shall learn how to find the sum of squares of first n natural numbers in Kotlin, using the formula or a loop statement, with example programs.

Kotlin – Sum of squares of first N natural numbers

To find the sum of squares of first n natural numbers in Kotlin, we can use the formula or a looping statement.

The formula to find the sum of squares of first n natural numbers is

n * ( n + 1 ) * ( 2 * n + 1) / 6

Program with Formula

In the following program, we read n from user and find the sum of squares of first n natural numbers using the specified formula.

Main.kt

fun main() {
    print("Enter n : ")
    val n = readLine()!!.toInt()

    val sum = n * (n + 1) * (2*n + 1) / 6
    print("Sum of squares of first $n natural numbers is $sum")
}

Output #1

Enter n : 6
Sum of squares of first 6 natural numbers is 91

Output #2

Enter n : 10
Sum of squares of first 10 natural numbers is 385

Related tutorials for the above program

• Kotlin Addition
• Kotlin Multiplication
• Kotlin Division

Program with Looping statement

In the following program, we read n from user and find the sum of first n natural numbers using For loop statement.

Main.kt

fun main() {
    print("Enter n : ")
    val n = readLine()!!.toInt()

    var sum = 0
    for (i in 1..n) {
        sum += i * i
    }
    print("Sum of squares of first $n natural numbers is $sum")
}

Output #1

Enter n : 6
Sum of squares of first 6 natural numbers is 91

Output #2

Enter n : 10
Sum of squares of first 10 natural numbers is 385

Related tutorials for the above program

• Kotlin – For i in range

Conclusion

In this Kotlin Tutorial, we learned how to find the sum of squares of first n natural numbers.