Python R4.2 Write a loop that computes a. The sum of all even numbers between 2

Question

Python R4.2 Write a loop that computes
a. The sum of all even numbers between 2 and 100 (inclusive).
b. The sum of all squares between 1 and 100 (inclusive).
c. The sum of all odd numbers between a and b (inclusive).
d. The sum of all odd digits of n. (For example, if n is 32677, the sum would be 3 + 7 + 7 = 17.)
Python R4.2 Write a loop that computes
a. The sum of all even numbers between 2 and 100 (inclusive).
b. The sum of all squares between 1 and 100 (inclusive).
c. The sum of all odd numbers between a and b (inclusive).
d. The sum of all odd digits of n. (For example, if n is 32677, the sum would be 3 + 7 + 7 = 17.)
R4.2 Write a loop that computes
a. The sum of all even numbers between 2 and 100 (inclusive).
b. The sum of all squares between 1 and 100 (inclusive).
c. The sum of all odd numbers between a and b (inclusive).
d. The sum of all odd digits of n. (For example, if n is 32677, the sum would be 3 + 7 + 7 = 17.)
R4.2 Write a loop that computes
a. The sum of all even numbers between 2 and 100 (inclusive).
b. The sum of all squares between 1 and 100 (inclusive).
c. The sum of all odd numbers between a and b (inclusive).
d. The sum of all odd digits of n. (For example, if n is 32677, the sum would be 3 + 7 + 7 = 17.)

Explanation / Answer

Part a)

sum=0

for i in range(2,101,2):

sum+=i

Part b)

sum=0

for i in range(1,11,1):

sum+=i*i

Part c)

def func(a,b):

sum=0

if (a%2==0):

a=a+1

if(b%2==1):

b=b+1

for i in range(a,b+1,2):

sum+=i

print(sum)

Part d)

def digitsum(n):

sum=0

while(n):

dig=n%10

if(dig%2==1):

sum+=dig

n=(n-dig)/10

print(sum)

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