29 Replies
@Apu
no approach in mind 💀
Note for OP
+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.Use sum of agp formula
To simplify ig.
Opt
I think that works
No wait
That's wrong
I'm dumb
Wouldn't that give binomial thingies as coeff?
d/dx of GP mb
Now expand maybe?
I really don't want to
i got n(n+2)(2n-1)/12 from my approach but that not even in options
ðŸ˜
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)
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
calculus ke alawa koi aur algebraic approach for sum of series i wrote above
Oh right this is more effecient ig.
how to find sum tho
I didn't understand your logic.
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
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.
Oh yeah that can do
sum nahi aa rha na par
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
Im getting A from the sum
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)
A na?
Yup
ohh got it
wait wait yeh kaise aaya
i got from my classic approach but multi wale se kaise
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
power of what is 2 ?
(1+x+2x^2+...) wala thing
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
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)
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.
ill manage with my classic approach 💀
thanx for the effort tho
+solved @Opt @SirLancelotDuLac
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