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From the textbook:

Suppose $(1+x+x^2+...+x^k)^n = a_0 + a_1x+a_2x^2 + ... + a_{kn}x^{kn}$.

Here is the question I'm working on:

Show that $a_0 + a_1 + a_2 + ... + a_{kn} = (k+1)^n$.

I know I need to use the Binomial Theorem in some way because I can expand $(k+1)^n$ this as

$(k+1)^n$ = ${n}\choose{0}$ $k^n$ + ${n}\choose{1}$ $k^{n-1}$ + ...

How can I use this expansion, if done correctly, to get started on the proof?

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    How are you defining $a_0,a_1,a_2,...$? – 2017-02-25
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    Ahh hold on I misread the question. I will add some more info now. – 2017-02-25
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    Plug $x=1$ into the equation. Done! – 2017-02-25
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    You're right! Hmm I'd thought it be a page-long proof. – 2017-02-25
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    Sorry,@quasi. I didn't notice your comment. – 2017-02-25

1 Answers 1

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You do not need the binomial theorem. Just put in $x=1$. Note that you get $$a_0 + a_1 + a_2 + ... + a_{kn} = (\underbrace{1+1+\dots+1}_{k+1 \textrm{ times}})^n=(k+1)^n$$

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    Yes that's true. quasi just pointed that out. Now it's clear. – 2017-02-25
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    @i8Σπ_821 So he did. – 2017-02-25
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    No prob -- nice to see it written out. – 2017-02-25
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    If only i'd known this sooner. Well part b of the proof is very similar to this one. – 2017-02-25
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    @i8Σπ_821 I suggest you remember the property "the sum of coefficients in a polynomial" is basically "plugging in $x=1$". Good luck! – 2017-02-25
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    I'll write that down as a reference for my exam. Thanks S.C.B and everyone! – 2017-02-25