C++ Program for Sum of squares of first n natural numbers Last Updated : 23 Jul, 2025 Comments Improve Suggest changes Like Article Like Report Given a positive integer N. The task is to find 12 + 22 + 32 + ..... + N2. Examples: Input : N = 4 Output : 30 12 + 22 + 32 + 42 = 1 + 4 + 9 + 16 = 30 Input : N = 5 Output : 55 Method 1: O(N) The idea is to run a loop from 1 to n and for each i, 1 <= i <= n, find i2 to sum. CPP // CPP Program to find sum of square of first n natural numbers #include <bits/stdc++.h> using namespace std; // Return the sum of the square // of first n natural numbers int squaresum(int n) { // Iterate i from 1 and n // finding square of i and add to sum. int sum = 0; for (int i = 1; i <= n; i++) sum += (i * i); return sum; } // Driven Program int main() { int n = 4; cout << squaresum(n) << endl; return 0; } Output: 30 Method 2: O(1) Proof: We know, (k + 1)3 = k3 + 3 * k2 + 3 * k + 1 We can write the above identity for k from 1 to n: 23 = 13 + 3 * 12 + 3 * 1 + 1 ......... (1) 33 = 23 + 3 * 22 + 3 * 2 + 1 ......... (2) 43 = 33 + 3 * 32 + 3 * 3 + 1 ......... (3) 53 = 43 + 3 * 42 + 3 * 4 + 1 ......... (4) ... n3 = (n - 1)3 + 3 * (n - 1)2 + 3 * (n - 1) + 1 ......... (n - 1) (n + 1)3 = n3 + 3 * n2 + 3 * n + 1 ......... (n) Putting equation (n - 1) in equation n, (n + 1)3 = (n - 1)3 + 3 * (n - 1)2 + 3 * (n - 1) + 1 + 3 * n2 + 3 * n + 1 = (n - 1)3 + 3 * (n2 + (n - 1)2) + 3 * ( n + (n - 1) ) + 1 + 1 By putting all equation, we get (n + 1)3 = 13 + 3 * ? k2 + 3 * ? k + ? 1 n3 + 3 * n2 + 3 * n + 1 = 1 + 3 * ? k2 + 3 * (n * (n + 1))/2 + n n3 + 3 * n2 + 3 * n = 3 * ? k2 + 3 * (n * (n + 1))/2 + n n3 + 3 * n2 + 2 * n - 3 * (n * (n + 1))/2 = 3 * ? k2 n * (n2 + 3 * n + 2) - 3 * (n * (n + 1))/2 = 3 * ? k2 n * (n + 1) * (n + 2) - 3 * (n * (n + 1))/2 = 3 * ? k2 n * (n + 1) * (n + 2 - 3/2) = 3 * ? k2 n * (n + 1) * (2 * n + 1)/2 = 3 * ? k2 n * (n + 1) * (2 * n + 1)/6 = ? k2 CPP // CPP Program to find sum // of square of first n // natural numbers #include <bits/stdc++.h> using namespace std; // Return the sum of square of // first n natural numbers int squaresum(int n) { return (n * (n + 1) * (2 * n + 1)) / 6; } // Driven Program int main() { int n = 4; cout << squaresum(n) << endl; return 0; } Output: 30 Avoiding early overflow: For large n, the value of (n * (n + 1) * (2 * n + 1)) would overflow. We can avoid overflow up to some extent using the fact that n*(n+1) must be divisible by 2. CPP // CPP Program to find sum of square of first // n natural numbers. This program avoids // overflow upto some extent for large value // of n. #include <bits/stdc++.h> using namespace std; // Return the sum of square of first n natural // numbers int squaresum(int n) { return (n * (n + 1) / 2) * (2 * n + 1) / 3; } // Driven Program int main() { int n = 4; cout << squaresum(n) << endl; return 0; } Output: 30 Please refer complete article on Sum of squares of first n natural numbers for more details! CPP Program for Sum of Squares of First 'n' Natural Numbers Comment More infoAdvertise with us Next Article Introduction to C++ Programming Language K kartik Follow Improve Article Tags : C++ Practice Tags : CPP Similar Reads C++ Programming Language C++ is a computer programming language developed by Bjarne Stroustrup as an extension of the C language. 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