y=[1:(2^x)];
y=sum(y);

great!!

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.

try: y=sum(1:2^x). It will lead to shorter solution.

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.