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Write a public method named findSum that takes a a parameter named n of type int. If n is less than 1 the method returns -1 otherwise it returns a number that is the result of applying the following formula to i values ranging from 1 to n: Note :E is the symbol for summation.

Write a public method named findSum that takes a a parameter named n of type int. If n is less than 1 the method returns -1 otherwise it returns a number that is the result of applying the following formula to i values ranging from 1 to n: Note :E is the symbol for summation.

Database System Concepts
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
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**Creating the `findSum` Method** **Objective:** Write a public method named `findSum` that takes a parameter named `n` of type `int`. If `n` is less than 1, the method returns -1; otherwise, it returns a number that is the result of applying the following formula to `i` values ranging from 1 to `n`. **Note:** The symbol Σ (summation) is used in the formula. **Formula Representation:** \[ \sum_{i=1}^{n} \left( \frac{n(n+1)}{2} \right) \] - **Explanation of the Formula:** - The formula provided within the summation symbol (Σ) calculates the sum of the first `n` natural numbers. - The term \(\frac{n(n+1)}{2}\) is derived from the arithmetic series sum formula for the first `n` natural numbers. Here is a breakdown of the formula: - **Summation (Σ):** This notation indicates that you sum the values of the given expression for `i` ranging from 1 to `n`. - **Expression:** \(\frac{n(n+1)}{2}\) - `n` is the given number. - The numerator `n(n+1)` calculates the product of `n` and `n+1`. - Dividing by 2 gives the sum of the first `n` natural numbers. **Instructions for Implementation:** - **Step 1:** First, check if `n` is less than 1. If so, return -1. - **Step 2:** Otherwise, calculate the sum using the formula \(\frac{n(n+1)}{2}\). - **Step 3:** Return the computed sum. By implementing these steps, the `findSum` method will accurately compute the required sum or return -1 if the input is invalid (i.e., less than 1).
Transcribed Image Text:**Creating the `findSum` Method** **Objective:** Write a public method named `findSum` that takes a parameter named `n` of type `int`. If `n` is less than 1, the method returns -1; otherwise, it returns a number that is the result of applying the following formula to `i` values ranging from 1 to `n`. **Note:** The symbol Σ (summation) is used in the formula. **Formula Representation:** \[ \sum_{i=1}^{n} \left( \frac{n(n+1)}{2} \right) \] - **Explanation of the Formula:** - The formula provided within the summation symbol (Σ) calculates the sum of the first `n` natural numbers. - The term \(\frac{n(n+1)}{2}\) is derived from the arithmetic series sum formula for the first `n` natural numbers. Here is a breakdown of the formula: - **Summation (Σ):** This notation indicates that you sum the values of the given expression for `i` ranging from 1 to `n`. - **Expression:** \(\frac{n(n+1)}{2}\) - `n` is the given number. - The numerator `n(n+1)` calculates the product of `n` and `n+1`. - Dividing by 2 gives the sum of the first `n` natural numbers. **Instructions for Implementation:** - **Step 1:** First, check if `n` is less than 1. If so, return -1. - **Step 2:** Otherwise, calculate the sum using the formula \(\frac{n(n+1)}{2}\). - **Step 3:** Return the computed sum. By implementing these steps, the `findSum` method will accurately compute the required sum or return -1 if the input is invalid (i.e., less than 1).
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