#Anyone knows the approach to this
misty field
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misty field
combinatorics
but that will make it undefined
for ncr to be defined, n has to be greater than or equal to r
and n cant be 0
it has to be any natural number
such that it would satisfy the sequence
rose viper
To start with, you remember the actual expression for nCk?
misty field
To start with, you remember the actual expression for `nCk`?
yes
rose viper
But what?
final cobalt
misty field
But what?
like
suppose you have to find the nth term
and n has to be any natural number
which year or grade?
oho
final year
XD
rose viper
and n has to be any natural number
The point is that the series is finite for any finite n.
Because for any finite n, at some point all terms go to 0.
misty field
Because for any finite n, at some point all terms go to 0.
yes
idk how to explain like you have to make a formula such that the thing will be defined
so something the would always make it defined
rose viper
It is always defined.
misty field
It *is* always defined.
yeah okay but then how do you find the general formula to it
rose viper
<:sobbingTEArs:1154982761925455932> yeah okay but then how do you find the gener...
...you start with the general formula for nCk.
misty field
i tried....
i got
wait
but it wouldnt make sense
like if you put 2
you would get 2c66
which is wrong
wait
what if i equate and make two equations with n and k?
with the conditions so that it satisfies
rustic spruce
Hm... Well, I guess we can use series multisection.
analog ferry
Maybe use the 66ᵗʰ root of 1 and binomial expansion.
misty field
Maybe use the 66ᵗʰ root of 1 and binomial expansion.
What do you mean 66th root of one?
misty field
Hm... Well, I guess we can use series multisection.
Let me try
misty field
What do you mean 66th root of one?
Do you mean iota?
analog ferry
In ℂ
rose viper
Show your work.
analog ferry
Maybe instead of 66 you could try the cases 2 and 3 to see what happens.
misty field
Show your work.
thats purely wrong 💀
bruh dude i actually just wrote that by the patern
but then i realized
that if n was any natural number it woudnt make sense like plug 2
2c66
thats just not defined
misty field
Maybe instead of 66 you could try the cases 2 and 3 to see what happens.
aight
analog ferry
I looked up at what a series multisection is and it is the same as what I had in mind.
misty field
i think i got the answer
thank you so much y'all
(1/66)\sum_{k=0}^{65}(1+e^2πi/66)^{n}
uh
can this be simplified further down?
or is it just fine here
fossil cedar
this exponent is wrong
ig you could consider this as summing the nth powers of the roots of (insert another poly I'll let you find)
then get a recursion with newton's
but that's not exactly simplifying
misty field
oh
ill see tomorrow after school
gotta wake in 4 hours 😔 and do a bit of differentiation and computers
analog ferry
Use (1-z)(1+z+z²+...+zⁿ)=1-zⁿ⁺¹ to find formulas on the series multisection.
misty field
analog ferry
Sorry I cannot delete the penultimate message, I meant
$$\sum_{k=0}^{65}(1+\omega^k)^n=\sum_{k=0}^{65}\sum_{p=0}^n\binom{n}{p}\omega^{pk}
=\sum_{p=0}^n\binom{n}{p}\sum_{k=0}^{65}\omega^{pk}$$
fresh tideBOT
mimshell
analog ferry
Indeed it delivers your formula.
Maybe I would write $1+\omega^k=\omega^\frac k2(\omega^{-\frac{k}2}+\omega^\frac k2)$ to get powers of $\omega$ and cos, and taking the real part would allow to find the formula in https://mathworld.wolfram.com/SeriesMultisection.html
fresh tideBOT
mimshell
