Considering Pascal's triangle, we know that the sum of two adjacent numbers is the number directly below them. Hence we can write the question as the sum of the numbers in the row above it:
Very elegant, nice work, my way was a little longer by considering expansions of (1+x)^2n and (1-x)^2nConsidering Pascal's triangle, we know that the sum of two adjacent numbers is the number directly below them. Hence we can write the question as the sum of the numbers in the row above it:
This is just the sum of the row of Pascal's triangle, which is
Err, not true. The last term on the LHS alone is greater than the RHS...A nice result I've proven
I considered the geometric series:Err, not true. The last term on the LHS alone is greater than the RHS...
:/ Can't believe I missed that, thanks for the heads up. I'll ask this question again once everyone has forgotten about it.Yes. Check the number of terms on the LHS, there are n+1, not n. A minor change needs to be made to your formula for the sum of a finite geometric series.
Note that the LHS is::/ Can't believe I missed that, thanks for the heads up. I'll ask this question again once everyone has forgotten about it.
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New result:
Consider them in groups of four. Now consider the general case:
(The signs alternate in pairs)
(Provide proof of your value, either with latex or a brief outline of steps taken)
Elegant, nice workConsider them in groups of four. Now consider the general case:
Now we will note that:
So the sum of each group of four is just two times the sum of that group. Hence if we were to do it to all groups of four up to 100, we would get two times the sum of all the numbers from 1 to 100:
This is just double the 100th triangular number, which is just
So the answer is
I meant, prove your value :sI believe him.