binomial 2

[sour gust]

<@&1227988399579730072>

[tough isle]

no approach in mind 💀

[sour gust]

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[dreamy bloom]

Use sum of agp formula

To simplify ig.

[tidal spire]

Opt

[dreamy jewel]

I think that works

No wait

That's wrong

I'm dumb

[dreamy bloom]

Wouldn't that give binomial thingies as coeff?

[dreamy jewel]

d/dx of GP mb

Now expand maybe?

I really don't want to

[tough isle]

i got n(n+2)(2n-1)/12 from my approach but that not even in options

😭

i simply observed that coeff must be
n(1)+ 1(n-1) +2(n-2) + 3(n-3).......2(n-2)+1(n-1)+n(1)

[dreamy jewel]

what?

Huh?

Oh wait I can simplify this so much

We can add 1 to GP without any changes

Because d/dx of 1 = 0

[tough isle]

calculus ke alawa koi aur algebraic approach for sum of series i wrote above

[dreamy bloom]

Oh right this is more effecient ig.

[tough isle]

how to find sum tho

[dreamy jewel]

I didn't understand your logic.

[tough isle]

i just wrote down the expression and observed in how many ways can i get x^n

x^0 * x^n
x^1 * x^n-1
x^2 * x^n-2
....
x^n-1x^1
x^n x^0

multiplied their coeffs and sum

[dreamy jewel]

But finding the sum is just 2n+∑k(n-k) in which you can split the summation into sum of integers and sum of integers squares.

[dreamy jewel]

Oh yeah that can do

[tough isle]

sum nahi aa rha na par

[tough isle]

somehow terms kholi toh dikha n+n+2n+3n..... -(1+4+9+16....)

1. theres two n
2. not able to grasp the end terms of the two series

[dreamy bloom]

Im getting A from the sum

[dreamy bloom]

The multinomial coeffecient must look like 2!/1!1!*(k^1)(n-k)^1 for k=1,2,...[n/2]
So the sum of these must look like sigma (k)(n-k) +2n (as coeffecient of x^0 is NOT 0)

[dreamy jewel]

A na?

[dreamy jewel]

Yup

[tough isle]

ohh got it

[tough isle]

wait wait yeh kaise aaya

i got from my classic approach but multi wale se kaise

[dreamy bloom]

Basically the power is 2 so if one term is of the type x^k, the other one must be x^(n-k). Now write the multinomial coeffecient for this and sum this over from k=0 to k=[n/2]. Now since we want to avoid [n/2], it is better to sum this up to n and then divide by 2.

But this method means coeffecient of x^0 is 0 which is not our case so add 2n to this too

[tough isle]

power of what is 2 ?

[dreamy bloom]

(1+x+2x^2+...) wala thing

[tough isle]

i dont understand multinomial coefficient kaise likha

from what i understand about multinomial
in (x+y+z)^n coefficient of x^a y^b z^c (where a+b+c=n) is n!/a!b!c!

can u explain along the lines of this

[dreamy bloom]

Ye. Here the value of n=2, so either two terms are 1 or one term is 2 and the rest are 0's. Now since both the terms must multiply upto x^n, you get the case when two terms are 1 and rest are zero for the two terms whose power is 1 as x^k and x^(n-k)

[dreamy bloom]

Now there is a slight conundrum when n is even and you can have a term with power 2 which is (n/2)x^(n/2). To avoid this thing, this is written.

[tough isle]

ill manage with my classic approach 💀

thanx for the effort tho

+solved @dreamy jewel @dreamy bloom