C++ Program - Calculate sum of Squares of Natural numbers
In Mathematics, the natural numbers are all positive numbers which is used for counting like 1, 2, 3, 4, and so on. The smallest natural number is 1.
Objective: Write a C++ program which returns sum of squares of natural numbers starting from 1 to given natural number n, (12 + 22 + 32 + ... + n2).
Method 1: Using while loop
The example below shows how to use while loop to calculate sum of squares of first n natural numbers.
#include <iostream> using namespace std; int main (){ int n = 10; int i = 1; int sum = 0; //calculating sum of squares from 1 to n while(i <= n) { sum += i*i; i++; } cout<<"Sum is: "<<sum; return 0; }
The above code will give the following output:
Sum is: 385
Method 2: Using for loop
The same can be achieved using for loop. Consider the example below:
#include <iostream> using namespace std; int main (){ int n = 10; int sum = 0; //calculating sum of squares from 1 to n for(int i = 1; i <= n; i++) sum += i*i; cout<<"Sum is: "<<sum; return 0; }
The above code will give the following output:
Sum is: 385
Method 3: Using Recursion
Similarly, recursion can be used to calculate the sum.
#include <iostream> using namespace std; //recursive function static int Sum(int n) { if(n == 1) return 1; else return (n*n + Sum(n-1)); } int main (){ cout<<"Sum of Squares of first 10 natural numbers: " <<Sum(10)<<endl; cout<<"Sum of Squares of first 20 natural numbers: " <<Sum(20)<<endl; return 0; }
The above code will give the following output:
Sum of Squares of first 10 natural numbers: 385 Sum of Squares of first 20 natural numbers: 2870
Method 4: Using Mathematical Formula
The sum of squares of first n natural numbers can be mathematically expressed as:
#include <iostream> using namespace std; int main (){ int n = 10; //calculating sum of squares from 1 to n int sum = n*(n+1)*(2*n+1)/6; cout<<"Sum is: "<<sum; return 0; }
The above code will give the following output:
Sum is: 385
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