#weird sum
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inland swan
it's not a series, it's a finite sum
spring gale
@void compass stop promoting
void compass
ok
inland swan
I can only see one conceivable way
$$ = \sum _{k=1}^n \frac{2^{k-1}k}{(k+1)!} - \sum _{k=1}^n \frac{2^{k-1}}{(k+1)!} $$
stoic cloakBOT
aL
inland swan
but i don't see what pairs of cancelling terms we can get here
you can prove the initial equality via induction, so it is true
@latent arch
verbal shuttleBOT
@errant lichen has given 1 rep to @inland swan
void compass
you dont need to use induction
look
(k+1)!=(k+1)k!
after that the sum will become \sum_{k=1}^{n}\frac{2^{k-1}}{k!}(\frac{k-1}{k+1})
Then use decomposition of fraction to the quotient you get
After decompse this sum as the difference of two sums
Then you can expand each sum till n
you get
inland swan
that's a nice one @void compass thanks
verbal shuttleBOT
@inland swan has given 1 rep to @void compass
inland swan
@latent arch
verbal shuttleBOT
@errant lichen has given 1 rep to @void compass
void compass
hahahahahahahahahahaha
inland swan
also, let people make help topics here so you can reply, don't do it over dms
that's bad practice
inland swan
🍺
void compass
hahahahahaha soda