THIS IS A PYTHON HOMEWORK ## Consider the Following Equations### Equations:\[\sum_{i=1}^{n} (3i - 2)^2 = \frac{1}{2}n(6n^2 - 3n - 1) \quad (32)\]\[\sum_{i=1}^{n} i^3 = \frac{1}{4}n^2(n + 1)^2 \quad (33)\]\[\cos(x) \cdot \cos(2x) \cdots \cos(2^{n-1}x) = \frac{\sin(2^n x)}{2^n \sin(x)} \quad (34)\]For the last expression, \( n = 1, 2, 3, \ldots \) and \(\sin(x) \neq 0\).### Some Helpful Background:Before we move ahead, note that equations 32, 33, and 34 are examples of analytical (closed-form) solutions. This means that you can either solve them via the formulas as shown or calculate each term individually, and both solutions will match. Isn’t it amazing that some people figured out that rather than calculating and adding each term in a loop, you can directly get the final result by plugging the values in the formula?While this is fascinating to some of us, analytical solutions are not available for all problems. These examples are provided to help check that both loop-based (calculating each term iteratively) and closed-form solutions match. Equations 35 to 37 demonstrate this for \( n = 5 \).### Loop-Based Calculation:\[\sum_{i=1}^{5} (3i - 2)^2 = (3 - 2)^2 + (6 - 2)^2 + (9 - 2)^2 + (12 - 2)^2 + (15 - 2)^2 \quad (35)\]Breaking it down:\[= 1^2 + 4^2 + 7^2 + 10^2 + 13^2 = 1 + 16 + 49 + 100 + 169 = 335 \quad (36)\]### Closed-Form Solution:\[= (1/2)5(6(5^2) - 3(5)) - 1 = (5 ```python4 def closed(n):5 return (1/4)*(n**2)*(n+1)**26 sum = 07 #complete bounded loop here8 return [sum,closed(stop)]9 10 def iv(stop):11 def closed(n):12 return (1/2)*n*(6*(n**2)-(3*n)-1)13 sum = 014 #complete bounded loop here15 return [sum, closed(stop)]16 17 def vi(x,stop):18 def closed(x,n):19 return math.sin(x+(2**n))/((2**n)*math.sin(x))20 prod = 121 #complete bounded loop here22 return [prod,closed(x,stop)]23 24 print(iii(5))25 print(iv(5))26 print(vi(math.pi,5))```**Explanation:**This Python code contains three functions: `closed`, `iv`, and `vi`, each designed to calculate a mathematical expression based on the given parameters and return a list of values. The functions `iv` and `vi` contain nested functions also named `closed`.- **Function `closed(n)`**: - Computes an expression using the formula \((1/4) \cdot (n^2) \cdot (n+1)^2\). - Returns a list containing the sum (uninitialized loop) and the result of the expression with `n` equal to `stop`.- **Function `iv(stop)`**: - Contains a nested function `closed(n)` which computes a different formula: \((1/2) \cdot n \cdot (6 \cdot (n^2) - (3 \cdot n) - 1)\). - Returns a list with the sum (uninitialized loop) and the result of the nested `closed(stop)`.- **Function `vi(x, stop)`**: - Contains a nested function `closed(x, n)` which calculates using trigonometric functions: \(\frac{\sin(x+(2^n))}{((2^n) \cdot \sin(x))}\). - Returns a list with the product (uninitialized loop) and the result of `closed(x, stop)`.Print statements show outputs

python introduction:- Python is a general-purpose, interpreted programming language. Python, which…

Consider the following equations. 5 Σ(3i – 2)2 i=1 = 1 import math 2 3 def iii(stop): = n = i=1 (3i - 2)² n i=1 23 cos(x) cos(2x)...cos(2n-1x) for the last expression n = 1, 2, 3, ... and sin(x) = 0. Some helpful background: Before we move ahead, note that equation-32, 33 and 34 are examples of analytical (closed from) solution. What it means is that, you can either solve it via the formula as shown in equations 32-34, or you can calculate each term individually, but the two solutions will match. Isn't that amazing that some people figured out that rather than calculating and adding each term in a loop, you can directly get the final result just by plugging the values in the formula :). Fasinating as it is to some of us, the analytical solutions are not available for all problems. We give them here for you to check that both loop based (going over each term) and the closed form solutions mathch. Equation-35 is the loop based calculation (for n = 5), and equation-37 is the direct result by plugging the value in the formula of equation-32. (6n²-3m = 3n-1) (n²(n + 1)² sin (2x) 2n sin(x) (3 − 2)² + (6 − 2)² + (9 − 2)² + (12 − 2)² + (15 − 2)² (32) 1² +4² +7² +10² + 13² = 1 + 16 +49 +100+ 169 = 335 (1/2)5(6(5²) – (3(5)) − 1) = (5/2)(150 – 16) = 335 (33) (34) (35) (36) (37) Complete the code below that uses a loop to calculate the sum (the closed-form is included to help you validate your code)

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

Problem 1PE