Question

help please 29

Answer 1

@Loser66

Answer 2

to me, the inside can be replace by
\[(tan^{-1}(2)-tan^{-1}(1))+tan^{-1}(3)-tan^{-1}(2))+\cdots\]

Answer 3

So, we cancel the like terms. At the end, we have \(-tan^{-1}(1) +tan^{-1}(\infty)\)

But \(tan^{-1}(1) = \pi/4 \) and \(tan^{-1}(\infty) = \pi/2\)

So the answer should be \(\pi/4\)

Let's wait for some others.

Answer 4

as we used to say "telescopes"

Answer 5

telescope series?

Answer 6

@Kainui

Answer 7

:d

Answer 8

Figure it out girl?

Answer 9

oh not yet D: i skipped it lol

Answer 10

@zepdrix come back online lmao

Answer 11

Did you not understand what Lose did?

Answer 12

no lol

Answer 13

:U

Answer 14

\[\rm \sum_{n=1}^{\infty}\left(Atan(n+1)-Atan(n)\right)\]Expand out the sum by inputting the n values,\[\rm =(Atan2-Atan1)+(Atan3-Atan2)+(Atan4-Atan3)+...\]

Answer 15

What you should notice is that a bunch of stuff is going to cancel out,\[\rm =(\cancel{Atan2}-Atan1)+(Atan3\cancel{-Atan2})+(Atan4-Atan3)+...\]because of all the subtraction, ya?

Answer 16

\[\rm =(\cancel{Atan2}-Atan1)+(\cancel{Atan3}\cancel{-Atan2})+(Atan4\cancel{-Atan3})+...\]

Answer 17

There is one term in the front that doesn't cancel out with anything.
And there will also be one term in the very end that doesn't cancel with anything.