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binomial theorem help

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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:

--- Quote from: sweet777 on November 02, 2009, 11:50:52 am ---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!!!!!!!!!!!!

--- End quote ---

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

Tr = nCr * yn-r * zr (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 * zr

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.

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