Question: # Block 1 computes the summation: Sum _ { j = 0 } ^ 1 7 ( 3 j + 2 ) sum = 0

# Block 1 computes the summation: Sum_{j =0}^17(3j +2)
sum =0
for j in range(0,18):
sum = sum +(3*j +2)
print('Sum_{j =0}^17(3j +2)='+ str(sum))
print('')
# Block 2 computes the summation: Sum_{j =0}^173j +2
sum =0
for j in range(0,18):
sum = sum +3*j
sum = sum +2
print('Sum_{j =0}^173j +2='+ str(sum))
print('')
# Block 3 computes the summation: Sum_{j =-2}^4(2j +1)^2
sum =0
# Add your code here
print('Sum_{j =-2}^4(2j +1)^2='+ str(sum))
print('')
# Block 4 computes the summation: Sum_{j =-1}^82^{j+4}+17
sum =0
# Add your code here
print('Sum_{j =-1}^82^{j+4}+17='+ str(sum))
print('')
# Block 5 computes the sum of the even integers between 5 and 17
sum =0
# Add your code here

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