#Summarize the sum???
deft helmBOT
paper bramble
is the answer 2021/2023?
mhm idk really
paper bramble
how did u find it?
dull solstice
u express 3 as n+2, like a general term
u feel me?
we basically use summation, taking the numerator as n+2 denominator as (n+2)! + (n+1)! + n!
taking n from 1 to 2018
paper bramble
u feel me?
yea
paper bramble
taking n from 1 to 2018
I understand
dull solstice
try and tell me if u still didnt get it
paper bramble
like this
paper bramble
I'm wrong ik ........
dull solstice
yess
paper bramble
but how to continue summarization?
paper bramble
do uk telescopic series
mhm yep
dull solstice
yes now think for a while how can u convert it
paper bramble
oke
dull solstice
got it?
dull solstice
can u tell what is the basic requirement for telescopic
paper bramble
serial limits??
dull solstice
umm, i mean like a (-) in the numerator lmao
paper bramble
mmm?
dull solstice
okay ill give the solution
paper bramble
yea
brave plover
I get $\frac{1}{2}-\frac{1}{2020!}$
dire tartanBOT
Jay
dark slate
Also ended up with this cool infinite series from adapting the problem.
brave plover
I got the same result, but as
\begin{align*} &\frac{2}{0!+1!+2!}+\frac{3}{1!+2!+3!}+\frac{4}{2!+3!+4!}+... \ &=\sum_{k=1}^{\infty}\frac{k+1}{(k-1)!+k!+(k+1)!} \ &=\lim_{n\to\infty}{\sum_{k=1}^n\frac{k+1}{(k-1)!+k!+(k+1)!}} \ &=\lim_{n\to\infty}{\sum_{k=1}^n\frac{k}{(k+1)!}} \ &=\lim_{n\to\infty}{\sum_{k=1}^n\left(\frac{1}{k!}-\frac{1}{(k+1)!}\right)} \ &=\lim_{n\to\infty}\left(1-\frac{1}{(n+1)!}\right) \&=1 \end{align*}
dire tartanBOT
Jay
dark slate
cool