The series to work on is as follows:
Illustration
Input : N = 4
Output: 30
Explanation: 12 + 22 + 32 + 42
= 1 + 4 + 9 + 16
= 30
Input: N = 5
Output: 55
Explanation: 12 + 22 + 32 + 42 + 52
= 1 + 4 + 9 + 16 + 25
= 55
Approaches: Here will be provided with the value up to which sum is computed for the series. in order to do so standard approaches are as follows:
- Naive Approach: Using for loop to compute the value
- Optimize Method: Using the formula to find the sum of the series.
Approach 1: Using loops a maintaining a count in a variable imposing condition inside the loop.
Example: Sum of the series by using for loop and calculate the square at each instance. This is the most naive approach.
Java
// Java Program to Add the nth Square Series// Importing java input output librariesimport java.io.*;class GFG { // Main driven Program public static void main(String args[]) throws IOException { // Declaring and initializing holding current sum int sum = 0; // Declaring VARIABLE holding term // Initializing to random value // to show output int n = 4; System.out.println("Enter number of terms:" + n); // For-loop for Iterating from 1 and Nth term for (int i = 1; i <= n; i++) { // finding square of current term and updating // current sum sum += (i * i); } // Printing final sum i.e // last updated current sum System.out.println("Sum for entered N terms:" + sum); }} |
Output:
30
Time Complexity: O(N)
Approach 2: In this approach, we will find the sum of the series by using the formula mentioned below. This formula can be used to find the sum of the nth Square Series and its correctness can be proved through mathematical induction. It is much more optimized the above illustrated.
12 + 22 + 32 + … + n2 = n * (n + 1) * (2 * (n + 1))/6
Example:
Java
// Java Program to Add the nth Square Series// Importing java input output librariesimport java.io.*;class GFG { // Main driven Program public static void main(String args[]) throws IOException { // Declaring VARIABLE holding term // Initializing to random value // to show output int n = 4; // Finding sum using formula int sum = (n * (n + 1) * (2 * n + 1)) / 6; // Displaying result System.out.println("Sum upto entered N series :"+sum); }} |
Output:
30
Time Complexity: O(1)
Auxiliary space: O(1) as it is using constant variables
