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What's the sum of coefficients of $(a+b+c)^8$?

Thanks in advance!

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If you're looking for a general way to do these types of expansion, you might want to look at this question. – Shaktal Apr 20 '14 at 19:03
up vote 6 down vote accepted

If you just imagine multiplying out the product using the distributive law, without collecting like terms, you end up with a bunch of terms like $acbbccab$ with $8$ factors each. Every 8-letter word built from the letters $a$, $b$, $c$ will appear exactly once, and there are $3^8$ such words.

When you then rearrange the sum by collecting like terms, the total sum of the coefficients will be exactly how many terms you had before you started collecting, that is, $3^8$. (This is because the coefficient of each of the terms is initially $1$).

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That will be $3^8 = 6561$. In general, if $f(x,y,z) = (x + y + z)^n$, then the sum of the coefficients is $f(1,1,1)$ which is $3^n$

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Why is that true? – charlie Apr 20 '14 at 9:39
Fill in $x = y = z = 1$ in (x + y + z)^n. Which is basically what LAcarguy said, only he used $f$ as a shorthand. – Ruben Apr 20 '14 at 9:41
No, I understood that, but why does that happen? – charlie Apr 20 '14 at 9:42
@charlie: if you set all the variables to $1$, the only thing that's left of each term is its coefficient, so the value of the entire expression is in that case the sum of the coefficients. – Henning Makholm Apr 20 '14 at 9:44

Taking the sum of coefficients just means setting all the variables equal to$~1$. So you've got $(1+1+1)^8=3^8=6561$.

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