...if you have the series 1+2+3...100 the answer is 5050. The formula for this is n(N+1)/2. Its cool. N stands for the number of terms in the sequence, which in this case, is 100.
It works for even numbers too: 2+4+6...+100. It works out to n(n+1).
Yes. Do you see why?
The sum of "the first n terms" is given by:
. . . . .\(\displaystyle 1\, +\, 2\, +\, 3\, +\, ...\, +\, n\, \)
. . . . . . .\(\displaystyle =\, 1\, +\, 2\, +\, 3\, +\, ...\, +\, 100\, \)
. . . . . . . . .\(\displaystyle =\, \dfrac{n\, (n\, +\, 1)}{2}\, =\, \dfrac{100\, (100\, +\, 1)}{2}\)
Since you are using only the even numbers, then the last term is not the n-th term, but the (n/2)-th term. In your first example, you added one hundred terms; in your second, you added only fifty. Since those fifty terms were even, they could be represented as:
. . . . .\(\displaystyle 2\, +\, 4\, +\, 6\, +\, ...\, +\, 2n\)
...where n = 50 (
not 100!). As a result, the summation would be from 1 through 50 (not 1 through 100), and thus the summation formulation would result in:
. . . . .\(\displaystyle 2\, +\, 4\, +\, 6\, +\, ...\, +\, 2n\, \)
. . . . . . .\(\displaystyle =\, 2\,(1\, +\, 2\, +\, 3\, +\, ...\, +\, 50)\, \)
. . . . . . . . .\(\displaystyle =\, 2\, \left(\dfrac{50\, (50\, +\, 1)}{2}\right)\, =\, 50\, (50\, +\, 1)\, =\, n\, (n\, +\, 1)\)
In this context, can 2+4+6...2n, be formulated as 2n(2n+1)? I would say no.
And you would be correct, as illustrated above.
If n is the number of terms in the sequence, what is 2n?
The value of "n" is the number of terms; the value of "2n" is the value in the summation. You cannot use "2n" as the counter (from the formula); in order to use the formula, you have to be careful and specific about the counter, the formula, and what you mean in each.
How do you explain that this is an illogical question?
I'm sorry, but I don't understand. What is the "illogical question"?
what math describes this?
What is the "this" that is being "described"? The summation formula is just algebra.
