2 Summations These three program snippets beep (by calling the beep() method). Looking at the code in each case you will see that the number of beeps depends only on the variable n. For each of the four a) write down a closed formula for the number of beeps using n (such as "3n+4" which is not one of the solutions) and b) write down how much that is for n = 1000. [3/each = 9]
2 Summations These three program snippets beep (by calling the beep() method). Looking at the code in each case you will see that the number of beeps depends only on the variable n. For each of the four a) write down a closed formula for the number of beeps using n (such as "3n+4" which is not one of the solutions) and b) write down how much that is for n = 1000. [3/each = 9]
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![2 Summations
These three program snippets beep (by calling the beep() method). Looking at the code
in each case you will see that the number of beeps depends only on the variable n.
For each of the four a) write down a closed formula for the number of beeps using n
(such as "3n +4" which is not one of the solutions) and b) write down how much that is
for n = 1000. [3/each = 9]
for (int i = 1; i <= 5; i++)
for (int j = 1; j <= n; j++)
beep();
for (int i = 1; i <= n; i++)
for (int j = 1; j <i; j++){
beep();
beep();
beep();
for (int i = 1; i<n; i++)
for (int j = 1; j<n; j+=n) {
beep0:
beep0:
beep();
number of beeps depending on n
number of beeps for n = 1000.
number of beeps depending on n
number of beeps for n = 1000
number of beeps depending on n
number of beeps for n = 1000](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F957a6242-1a60-46cb-823f-4a25075634cc%2Fc12622fb-8e93-4b20-9edc-a4eeb23c75b0%2F92oun7_processed.jpeg&w=3840&q=75)
Transcribed Image Text:2 Summations
These three program snippets beep (by calling the beep() method). Looking at the code
in each case you will see that the number of beeps depends only on the variable n.
For each of the four a) write down a closed formula for the number of beeps using n
(such as "3n +4" which is not one of the solutions) and b) write down how much that is
for n = 1000. [3/each = 9]
for (int i = 1; i <= 5; i++)
for (int j = 1; j <= n; j++)
beep();
for (int i = 1; i <= n; i++)
for (int j = 1; j <i; j++){
beep();
beep();
beep();
for (int i = 1; i<n; i++)
for (int j = 1; j<n; j+=n) {
beep0:
beep0:
beep();
number of beeps depending on n
number of beeps for n = 1000.
number of beeps depending on n
number of beeps for n = 1000
number of beeps depending on n
number of beeps for n = 1000
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