1. ## polya strikes out

Write down the numbers 1,2,3,4,... in a row Srike out every third
so
we are left with 1,2,4,5,7,8,10... now write their cumulative sums
of what remains. We are given
1,3,7,12,19,27,37... Now strike out every second
term then add the sums. So we are left with 8,27,64,125,216,...
The pattern is 2^3,3^3,4^3,5^3,...

How can I explain how this happens? I believe I have to write a formula for each step that I do. I do not know if this is right or how I could get the formula. Any help would be very nice. Thank you for your time.

2. If your previous post has not been answered it is either because no
one knows how to answer it. Posting it again does not help, perhaps
if you posted some of your thoughts in the original thread you would
get more help.

RonL

3. Originally Posted by schinb64
Write down the numbers 1,2,3,4,... in a row Srike out every third
so
we are left with 1,2,4,5,7,8,10... now write their cumulative sums
of what remains. We are given
1,3,7,12,19,27,37... Now strike out every second
term then add the sums. So we are left with 8,27,64,125,216,...
The pattern is 2^3,3^3,4^3,5^3,...

How can I explain how this happens? I believe I have to write a formula for each step that I do. I do not know if this is right or how I could get the formula. Any help would be very nice. Thank you for your time.
It seems like a long problem.

But it does help if you have a formula.

Let {b_n} be the sequence the represents after striking out every third one. Then, b_n=n+[(n-1)/2].
Where [ ] means "greatest integer function".

Okay, the next step is to find the sum of all terms before.

Thus,

SUM(k=1,n) b_k = SUM(k=1,n) k+[(k-1)/2]

To find this use the fact that,

SUM(k=1,n) k = n(n+1)/2

And

SUM(k=1,n) [k/2] = n^2 if n is even, otherwise n^2+n.