Problem 189. Sum all integers from 1 to 2^n

good excersize.

This solution accepts arrays and also leverages the fact that we need to sum to a power of two.

Great Question to learn for syntax

Can't believe someone flagged this.

nice problem

great!!

good problem

great!!!!!

great cheat!!!!

Very nice problem for kids and adults alike.

Celentano's solution is efficacious.

That was fun :)

nice

(^-^)V

any math formula?

:)

This user (mohamed elbesealy) appears to have gamed the system, with fraudulent "likes" of this unremarkable solution submitted by 'sock-puppet' accounts. —DIV

hint:
sum of sequential series
1+2+.....n= n(n+1)/2

Of course, this solution use the formula from Carl Friedrich Gauss for the sum of the integers from 0 to n.

While the obvious solution is y = sum(1:2^x), that will fail miserably for x = 50. So the alternative is a looping solution, that generates the sum more intelligently. Here, the looping is done simply using recursion. In fact, we can even compute the exact sum for x =100, a problem that would take the brute force solution the lifetime of the universe.

sum_int(sym(100))
ans =
803469022129495137770981046171215126561215611592144769253376

This done in fractions of a second, even for symbolic inputs.

not sure why this is wrong.

Upper limit must be 2^n, according to the problem's title & Test Suite. (Not clear from the problem statement, though!)

I can't figure this out after 10+ tries. Someone please help? http://www.followthesteps.net/sky-contact-phone-number/

Colon notation is key.

Ha!

nice

ha ha ha

I can't figure this out after 10+ tries. Someone please help?

http://www.mathworks.com/help/matlab/matlab_prog/create-functions-in-files.html

How this solution is possible?

Ha ha !

This solution can accept arrays of integers as x input

This solution accepts arrays, avoids the explicite creation of 1:2^n and calculates the expensive power once only.

A good example of a solution that does well on Cody's size measure, but which I wouldn't use for serious purposes.