binomial theorem help

[sweet777]

um how would you solve this question'?

in the expansion of (1+ax)² the first term is 1, 2nd term = 24x, 3rd term = 252x². find the value for a and n...

please tell me fast how u solve this...thanks!!!!!!!!!!!!

[sweet777]

thanks for the effort...but actually, the answer at d back of the book is a = 3 and n = 8...anyways...thanks!! thank you soo much!

[Ghost Of Highbury]

um how would you solve this question'?

in the expansion of (1+ax)² the first term is 1, 2nd term = 24x, 3rd term = 252x². find the value for a and n...

please tell me fast how u solve this...thanks!!!!!!!!!!!!

okk...first the question u wrote is wrong..

it shud be (1+ax)ⁿ

got the answer ..will reply in my next post..plzz wait..

[Ghost Of Highbury]

(1+ax)ⁿ

T_r = nCr * y^n-r * z^r (y+z)ⁿ...here y=1....z = ax

ok..

y i.e 1 raised to anything willl give u 1..so we omit that..

that leaves us with

nCr * z^r

when r = 1 the coefficient is 24

so..

nC1 * ax = 24x

n * a = 24

an = 24

a = 24/n

-----------

when r = 2 the coefficient is 252

nC2 * (ax)²

nC2 * a²x² = 252x²

nC2 = (n² - n)/2

= 252

a = 24/n
so, a² = 576/ⁿ²

substituting this we get

= 252


576n² - 576n
-----------------------  =  252
          2n²          


576n² - 576n = 504²

576n² - 504² = 576n

72n² = 576n

n = 8

substituting again in a = 24/n we get a = 3

[IGSTUDENT]

I need help to evaluate:

(1+ sq.root 3i) raised to 3.

Please help .I have a mock coming.

[sweet777]

"when r = 2 the coefficient is 252

nC2 * (ax)²

nC2 * a²x² = 252x²

nC2 = (n2 - n)/2"

hey..i gt dpart in which a=24/n....but i really dont understand how you got (n² - n) from a²x²=252x²