CODEWITHSUNDEEP
pythonmathsumsequenceseries
Cloud
Cloud

Reputation: 169

Sum a sequence of numbers

Given: Two positive integers a and b (a

Return: The sum of all odd integers from a through b, inclusively.

    #My code: 
    a = 100
    b = 200
    for i in range(a,b):
        if i%2 == 1:
            print i

At the moment it's just showing a drop down list of all the odd integers. I do not know how to affix a "range" to this properly, if need be. How can I add on to my code above to get the sum of all the odd integers?

Thanks

Upvotes: 1

Views: 22598

Answers (5)

kojiro
kojiro

Reputation: 77107

There are a bunch of ways to do this. If you think about the math, though, it's a lot like Gauss's old problem. Gauss was asked to add the numbers between 1 and 100, and he realized that each pair of high and low values summed to 101 (100 + 1, 99 + 2, 98 + 3…)

high = b
low = a

So we have to multiply some number of b + a values. How many are there? For all the integers, that's just

num_pairs = (high-low) // 2

Then we multiply that number by high + low to get the answer:

result = (high + low) * num_pairs

But you only want every other ones, so we divide by two again:

result //= 2

Totally:

def sumrange(low, high, step):
     num_pairs = (high - low) // 2
     result = (high + low) * num_pairs
     return result // step

or sumrange = lambda low, high, step: (high - low) * (high + low) // (2 * step)

Now this still isn't quite an answer to your question, because it needs to be offset depending on whether your low value is odd, and whether your high value is included or excluded. But I will leave that as an exercise.

Making this a CW answer so someone can edit if my math is messy.

Upvotes: 1

Yohst
Yohst

Reputation: 1902

And the numpy version of the solution:

import numpy as np
a = 100
b = 200
r = np.linspace(a,b-1,b-a)
r = np.sum(np.mod(r,2)*r)
print(r)

Upvotes: 0

Jagadeesh Pulamarasetti
Jagadeesh Pulamarasetti

Reputation: 483

Some mathematical tricks can solve your problem much efficiently.

For example, sum of first n odd numbers = n*n square(n)

So you can use for

Sum of odd numbers [m,n] = n*n - (m-2)*(m-2) where m!=1 and m and n are odds

One more useful analysis is, AP (arithmetic progression)

Formula : (n/2)*(a+l) where n= no. of elements, a = first term, l= last term

Here,

a = m [if m is odd]

a = m+1 [if m is even]

l = n [if n is odd]

l = n-1 [if n is even]

n = ( ( l - a ) / 2 ) + 1

By applying in code, you can easily get the answer...

Upvotes: 0

user278064
user278064

Reputation: 10170

Sum all the numbers between a and b if odd.

sum(i for i in xrange(a, b) if i%2)

Upvotes: 4

Ignacio Vergara Kausel
Ignacio Vergara Kausel

Reputation: 5996

A rather quick way to do it would be:

result = 0
for i in range(a,b+1):
  if i%2 == 1:
    result += i
print result

Upvotes: 3

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