Find the coefficient of x^{21} without calculating the product

[mathmari]

Hey!

"Let the polynomials
$$f(x)=1+\sum_{k=1}^{8}{(2k)x^{2k}} \text{ and } g(x)=1+ \sum_{k=1}^{8}{(3k)x^{3k}}$$
of $\mathbb{Q}$. Without calculating $f(x)g(x)$, find the coefficient of $x^{21}$ at $f(x)g(x)$."
Let's consider $f(x)=\sum_{k=0}^{\infty}{a_kx^k}$ and $g(x)=\sum_{k=0}^{\infty}{b_kx^k}$

So the following coefficients exist:
\begin{matrix}
a_0 & b_0 \\
a_2 & b_3 \\
a_4 & b_6 \\
a_6 & b_9 \\
a_8 & b_{12} \\
a_{10} & b_{15} \\
a_{12} & b_{18} \\
a_{14} & b_{21} \\
a_{16} & b_{24} \\
\end{matrix}

Since $c_{21}=\sum_{i+j=21}{a_i b_j}$, we have to find each time two coefficients for which the sum of their indices is equal to $21$. So:
$a_0, b_{21}$
$a_6, b_{15}$
$a_{12}, b_9$

Therefore, $c_8=a_0 b_{21}+a_6 b_{15}+ a_{12} b_9=21+6 \cdot 15+ 12 \cdot 9=219$.

Is this correct?
Is the way I solved it ok, or is there a better way to write it?

 

[Deveno]

That's fine.

What you are doing is solving:

$k = 0\text{ (mod }2)$
$21 - k = 0\text{ (mod }3)$

The second equation is clearly the same as:

$k = 0\text{ (mod }3)$ so together we get:

$k = 0\text{ (mod }6)$

Then we just find these values of $k$ with $0 \leq k \leq 16$ (since that's as high as the non-zero a's go).

*****

You should probably have used another index besides $k$, in your second formulation of $f$ and $g$. What you are interested in is the non-zero terms (technically ALL the terms "exist").

 

[mathmari]

That's fine.

What you are doing is solving:

$k = 0\text{ (mod }2)$
$21 - k = 0\text{ (mod }3)$

The second equation is clearly the same as:

$k = 0\text{ (mod }3)$ so together we get:

$k = 0\text{ (mod }6)$

Then we just find these values of $k$ with $0 \leq k \leq 16$ (since that's as high as the non-zero a's go).

*****

You should probably have used another index besides $k$, in your second formulation of $f$ and $g$. What you are interested in is the non-zero terms (technically ALL the terms "exist").

Aha! Ok!
So is the way I formulated the solution nice? Or is there a better way to express it?

 

[Deveno]

What you have done is fine, and correct.

 

[mathmari]

What you have done is fine, and correct.

Great! Thanks a lot! (Smirk)