How to find sum of the series 2+ 10+ 18+.... n in Python?

Question

I'm a new learner and wondering how to do it. I can do it in c programming

Answer 1

You just need to use range() and sum() :

result = sum(range(2, n, 8))

where n is the ending value. In case you don't know, the third argument of range() is optional (it defaults to 1) and it is the step used to output values from 2 to n, giving you 2, 2+8, 2+8+8,..., until the last 2+8k that is smaller than n.

Happy coding!

Answer 2

Gonçalo Magalhaes thank you so much man ❤️

Answer 3

You're welcome! Glad I could help :)

Answer 4

Ya you had i better answer. Didn't think of that.

Answer 5

The easiest solution is this:

num_list=[2, 4,3,5,1]
sum(num_list)
# returns sum of the number elements of the num_list.

Answer 6

👻 ƒαянαn 👻 {Less Active: exams} thanks man

Answer 7

Unless Python optimizes for this, it might be faster computationally to use the mathematical shortcut for an arithmetic series Sq= q*(a1 + aq) /2. In this case the nth term, n = 2+(q-1)*8.

So with the correction from thoq! below,
result=((n + 4)**2 - 4) /16

Here's code showing equivalence to sum(range(2,1000,8)), up to n=994. I haven't timed the two methods yet:
https://code.sololearn.com/c9XMNrJqtmRf/?ref=app

Or in terms of the qth term:
result= (4 *q**2)-2*q

I'll time this at some point to see if there's any speed difference compared to writing out the sum of a range. I'm guessing there won't be much.

https://m.wikihow.com/Find-the-Sum-of-an-Arithmetic-Sequence

Answer 8

Hope J

Your formula doesn't seem to work. This one does (I am sure it could be simplified):

res = 8*(((n-2)/8 + 1)*(n-2)/8/2)+2*((n-2)/8)+2

Gonçalo Magalhaes

It needs to be like this:

sum(range(2,n+1,8))

Otherwise n is excluded from the sum.

Answer 9

x=0
n=int(input("Enter the number"))
for a in range(2,n+1,8):
x=x+a
print(x)

Answer 10

If it's a string you can use the eval() function

Answer 11

not string, i want the sum

Answer 12

I will not give you the exact code.
What you do is
1. Define an empty list and the variable, say x, with value 2
2. While loop till x is not equal to n, adding 8 to x each time.
3. In each loop, add the new value of x to the list
4. In the end, find the sum of the list

Hope I was clear and also that i understood what you want correctly

Answer 13

Thank you very much XXX, but could you please write the code?

Answer 14

https://code.sololearn.com/clL46z1HnD25/?ref=app

hope that helps.
Just a suggestion, you should try yourself also

Answer 15

It works for me. n is the number of terms in the series (including the initial '2'). a(n) is the n'th term, s(n) is the sum of the first n terms.

The formula I gave simplifies to: 4n^2-2n.

n=1: 2 = 2.
s(n)=4(1^2)-2(1) = 4-2= 2.

n=2: 2 + 10 = 12.
s(n)=4(2^2)-2(2)= 16 - 4 =12.

n=3: 2 + 10 + 18 = 30.
s(n)=4(3^2)-2(3)=4(9)-6=30.

n=4: 2+10+18+26=56.
s(n)=4(4^2)-2(4)=56.

more generally, if you assume s(n) = 4n^2 - 2n, and we know that that the nth term a(n) = 2 + 8(n-1),
you can show that s(n+1)=s(n)+a(n+1), so qed by induction.

Answer 16

Hope J I just tried your formula. Now that I am looking more closely at it I can see that it doesn't work because you were mixing up the variables as they were given in the task. There n is the last number in the series and not the count of numbers in the series.

Answer 17

Ok right. I think the right formula then is s(n) = [(n + 4)^2 - 4]/16.

I only verified up to n=26 (four terms) .