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Binomial Theorem (P1) - doubt

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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  :o

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  ;D

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