Please can be handwritten. Question 2: Implementing a Recursive Function . Write recursive function, recursionprob(n), which takes a positive number as its argument and returns the output as shown below.The solution should clearly write the steps as shown in an example in slide number 59 and slide number 60 in lecture slides.After writing the steps, trace the function for “recursiveprob(5)” as shown in an example slide number 61.Function Output: >> recursionprob(1) 1 >> recursionprob(2) 1 4 >> recursionprob(3) 1 4 9 >>recrusionprob(4) 1 4 9 16 Solve Using Recursion• The problem is to find the sum for a given vale of 'n' such that:• sum=1+2+.....+nWhen n is 1, the answer is 1, when n is 2, the answer is 1+2= 3, when n is 3, the answer is 1+2+3 =6, when n is 4, the answer is 1+2+3+4= 10 and goes on....Using Recursion:Step 1: The base cases is when n is 1, the result is 1.Step 2: Base Case: if (n == 1), return 1.• Step 3: Function implementation for Base Case:• int sum(n){if(n== 1).(return 1:1)Step 4: Using Recursion:•When n is 2, it is equal to sum of when ((n is 1) + 2). It is (n-1) + n when n is 2.•When n is 3, it is equal to sum of ((n is 1)+(n is 2)+3). So, it is (n-2)+(n-1)+ n when n is 3.When n is 4, it is equal to sum of ((n is 1)+(n is 2)+(n is 3) + 4). So, it is (n-3)+(n-2)+(n-1) + n.• The repetitive logic is adding n with previous value of sum of (n-1).Output is returned as 10Trace the RecursionStep 8: return 10Step 7: return 6sum(4)Step 1: execute sum(4)return sum(3) + 41Step 6: return 359Step 5: return 1Using Recursion:Step 5: The recursive logic is:Step 2: execute sum(3)return sum(2) + 3 TStep 6: Recursion Implementation:int sum(n)1• when (n greater than 1)→ sum(n-1)+nStep 3: execute sum(2)return sum(1) +2Solve Using Recursionreturn 1if(n== 1)11else111return 1:return sum(n-1) + mStep 4: execute sum(1)161

Step 1 : STARTStep 2 : implement the recursionprob functionStep 3 : call the function with parameter…

Answered: Please can be handwritten. Question 2:…

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

Problem 1PE

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Please can be handwritten. Question 2: Implementing a Recursive Function . Write recursive function, recursionprob(n), which takes a positive number as its argument and returns the output as shown below.

The solution should clearly write the steps as shown in an example in slide number 59 and slide number 60 in lecture slides.

After writing the steps, trace the function for “recursiveprob(5)” as shown in an example slide number 61.

Function Output: >> recursionprob(1) 1 >> recursionprob(2) 1 4 >> recursionprob(3) 1 4 9 >>recrusionprob(4) 1 4 9 16

Solve Using Recursion
• The problem is to find the sum for a given vale of 'n' such that:
• sum=1+2+.....+n
When n is 1, the answer is 1, when n is 2, the answer is 1+2= 3, when n is 3, the answer is 1+2+3 =
6, when n is 4, the answer is 1+2+3+4= 10 and goes on....
Using Recursion:
Step 1: The base cases is when n is 1, the result is 1.
Step 2: Base Case: if (n == 1), return 1.
• Step 3: Function implementation for Base Case:
• int sum(n){
if(n== 1)
.
(return 1:1)
Step 4: Using Recursion:
•When n is 2, it is equal to sum of when ((n is 1) + 2). It is (n-1) + n when n is 2.
•When n is 3, it is equal to sum of ((n is 1)+(n is 2)+3). So, it is (n-2)+(n-1)+ n when n is 3.
When n is 4, it is equal to sum of ((n is 1)+(n is 2)+(n is 3) + 4). So, it is (n-3)+(n-2)+(n-1) + n.
• The repetitive logic is adding n with previous value of sum of (n-1).
Output is returned as 10
Trace the Recursion
Step 8: return 10
Step 7: return 6
sum(4)
Step 1: execute sum(4)
return sum(3) + 4
1
Step 6: return 3
59
Step 5: return 1
Using Recursion:
Step 5: The recursive logic is:
Step 2: execute sum(3)
return sum(2) + 3 T
Step 6: Recursion Implementation:
int sum(n)
1
• when (n greater than 1)→ sum(n-1)+n
Step 3: execute sum(2)
return sum(1) +2
Solve Using Recursion
return 1
if(n== 1)
1
1
else
1
1
1
return 1:
return sum(n-1) + m
Step 4: execute sum(1)
1
61

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