Binomial Theorem (P1) - doubt

[EMO123]

Hey everyone. I am new here.

Doubt is :

Given that the expansion of (1 + ax)ⁿ begins 1 + 36x + 576x², find the values of a and n.

Thank you.

[EMO123]

Is there anyone who can solve my doubt 

[Khey [Rainbow]]

Hey EMO123

Here is the solution to ur query:
(1+ax)^n = 1+36x+576x^2
1+n(ax)+[n(n-1)(ax)^2]/2! = 1+36x+576x^2

na=36
n= 36/a  

subsitute 36/a for n
[36/a(36/a-1)a^2] divide by 2 factorial and this equals to 576
and u'll end up with an equation
(1296a-36a^2)/a=1152
36a^2+1152a-1296a=0
a(36a-144)=0
36a=144
a=4
n=36/4
n=9

HOPE THIS HELPS 

[EMO123]

thanxx khey for your help

[Khey [Rainbow]]

Anytym 

[EMO123]

there is another question posted can u plz solve that to
thanxx for the ans

[astarmathsandphysics]

where is this other question

[Khey [Rainbow]]

Emo123
where is the other question???

[elemis]

Emo123
where is the other question???

Its been solved.

[EMO123]

it is a new topic