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Author Topic: binomial theorem help  (Read 285 times)
sweet777
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« 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!!!!!!!!!!!!
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sweet777
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« Reply #1 on: November 02, 2009, 03:06:24 PM »

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


« Reply #2 on: November 02, 2009, 03:12:51 PM »

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..
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« Reply #3 on: November 02, 2009, 03:22:48 PM »

(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

(n^2 - n)/2 *a^2 = 252

a = 24/n
so, a2 = 576/n2

substituting this we get

(n^2 - n)/2 *(576/n^2) = 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
« Last Edit: November 02, 2009, 03:24:58 PM by A@di » Logged

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« Reply #4 on: November 15, 2009, 02:40:58 AM »

I need help to evaluate:

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

Please help .I have a mock coming.

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« Reply #5 on: November 20, 2009, 12:52:53 AM »

"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
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