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Sunday, November 2, 2008

Q) Find the coefficient of x^4 in (1 + 2x + 3x^2)^5 .

A) [(1+2x)+3x^2)]^5

'r'th Term = 5Cr (1+2x)^r . (3x^2)^{5-r}

To form net effective 4th powers of x:
from the righthand term, we can only use x^0 x^2 and x^4
so from the lefthand term (1+2x)^r ... we use x^4, x^2 and x^0 respectively.

5C5.(3x^2)^0 . (1+2x)^5 -----------> 5C5. 5C4.(2x)^4
and
5C4.(3x^2).(1+2x)^4 --------> 5C4.(3x^2).[4C2.(2x)^2]
and
5C3.(3x^2)^2 .(1+2x)^3 ---------> 5C3.(3x^2)^2 [3C0. (2x)^0]

So, coefficient of x^4 :

5C5.5C4.2^4 + 5C4.3.4C2. 2^2 + 5C3.3^2. 1
= 5. 16 + 5.3.6.4 + 10.9
= 80+ 360 + 90
= 530

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2 comments:

Ashwin Ravichandran said...

Hmm. A quicker method would be the multinomial theorem, bro.

Find the coefficient of x^4 in (1 + 2x + 3x^2)^5 .
> p + q + r = 5

> [5!/[p!*q!*r!]*(1^p)*(2^q)*(x^q)*(3^r)*(x^2r)

> Basically the powers of the x - terms have to be equal to 4.
q + 2r = 4
By elementary:
We have 3 cases,
(4,0) p = 1
(2,1) p = 2
(0,2) p = 3

Apply it in the above expression:
80 + 360 + 90 = 530

spideyunlimited said...

Good work Ashwin, that's how I confirmed my answer later.
However, on this site I post the solutions I gave on GoIIT as is; and there I just did whatever struck me then and there... the 1 term made it easy 'cos of x^0 all along.