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- Feb 14, 2012

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Hi,

I want to find the coefficient of $\displaystyle x^4 $ and $\displaystyle x^{39} $ of $\displaystyle (1+x+x^2)^{20} $

I use the binomial theorem twice and come out with the following argument:

$\displaystyle (1+x+2x^2)^{20}= [(1+x)+(2x^2)]^{20}$

I see that

$\displaystyle T(r+1)={20 \choose r} (1+x)^{20-r}(2x^2)^r $

$\displaystyle T(r+1)={20 \choose r} .2^r. x^{2r}.(1+x)^{20-r} $

$\displaystyle T(r+1)={20 \choose r} .2^r. x^{2r}.{20-r \choose k} (1)^{20-r-k}. (x)^{k}$

$\displaystyle T(r+1)={20 \choose r} {20-r \choose k} .2^r. x^{2r+k}$

Now, I notice that $\displaystyle 20-r \ge k$, and, of course, $\displaystyle 0\le r,k \le 20 $.

To obtain the coefficient of x^4, we need 2r+k=4

If k=0, r=2. (Case 1)...this gives the coefficient of 760.

If k=1, then r is not an integer.

If k=2, r=1 (Case 2)...this gives the coefficient of 6840.

If k=4, r=0 (Case 3)...this gives the coefficient of 4845.

Hence, the coefficient of x^4 is (760+6840+4845=12445).

And I do the same for x^39. It turns out that there is only one case to generate the term x^39, i.e. when k=1, r=19. Thus the coefficient of x^39 is 20(2^19).

All this is only common sense and it certainly is not an elegant method to find for coefficient of any specific term. I wish to ask if there exists a general formula to find the coefficient of trinomial expansion of the type (a+bx+cx^2)?

Thanks.

I want to find the coefficient of $\displaystyle x^4 $ and $\displaystyle x^{39} $ of $\displaystyle (1+x+x^2)^{20} $

I use the binomial theorem twice and come out with the following argument:

$\displaystyle (1+x+2x^2)^{20}= [(1+x)+(2x^2)]^{20}$

I see that

$\displaystyle T(r+1)={20 \choose r} (1+x)^{20-r}(2x^2)^r $

$\displaystyle T(r+1)={20 \choose r} .2^r. x^{2r}.(1+x)^{20-r} $

$\displaystyle T(r+1)={20 \choose r} .2^r. x^{2r}.{20-r \choose k} (1)^{20-r-k}. (x)^{k}$

$\displaystyle T(r+1)={20 \choose r} {20-r \choose k} .2^r. x^{2r+k}$

Now, I notice that $\displaystyle 20-r \ge k$, and, of course, $\displaystyle 0\le r,k \le 20 $.

To obtain the coefficient of x^4, we need 2r+k=4

If k=0, r=2. (Case 1)...this gives the coefficient of 760.

If k=1, then r is not an integer.

If k=2, r=1 (Case 2)...this gives the coefficient of 6840.

If k=4, r=0 (Case 3)...this gives the coefficient of 4845.

Hence, the coefficient of x^4 is (760+6840+4845=12445).

And I do the same for x^39. It turns out that there is only one case to generate the term x^39, i.e. when k=1, r=19. Thus the coefficient of x^39 is 20(2^19).

All this is only common sense and it certainly is not an elegant method to find for coefficient of any specific term. I wish to ask if there exists a general formula to find the coefficient of trinomial expansion of the type (a+bx+cx^2)?

Thanks.

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