<s>[INST] Need an explanation for binomial theorem question I understand the basic concept of binomial theorem, however once I begin looking at this question I can't seem to wrap my head around why x^0 is equivalent x^16 once you change the form of the question (This is off a short revision quiz). If anyone could give me a quick rundown on how to solve this question or even link me to a resource where I could figure it out, It would be much appreciated!
Thanks!



*Find the coefficient of $x^0$ in $(2x^2+5x^{-2})^8$.

Solution: Rewriting $(2x^2+4x^{-2})^8$ as $(x^{-2})^8(2x^4+5)^8$, we see that finding the coefficient of $x^0$ is equivalent to finding the coefficient of $x^{16}$ in $$(2x^4+5)^8=\sum_{i=0}^8\binom8{8-i}(2x^4)^{8-i}5^i\;.$$ Now $16=4(8-i)$, so $i=4$. Hence the coefficient is $$\binom842^45^4=\frac{8!}{4!4!}2^45^4=70\times 16\times 625\;.$$

 [/INST] I’m assuming that you have no problem with the fact that $(2x^2+5x^{-2})^8=(x^{-2})^8(2x^4+5)^8$, but just in case: $2x^2+5x^{-2}=x^{-2}(2x^4+5)$.
We want the coefficient of $x^0$ in $(2x^2+5x^{-2})^8$; this is of course the same as the coefficient of $x^0$ in $(x^{-2})^8(2x^4+5)^8$. Note that $(x^{-2})^8(2x^4+5)^8=x^{-16}(2x^4+5)^8$, and suppose that we’ve multiplied out $(2x^4+5)^8$ and got some polynomial with terms like $ax^k$. When we finish the calculation by multiplying this polynomial by $x^{-16}$, each of its terms $ax^k$ will become $ax^{k-16}$. We want the term with exponent $0$ in the final product, which we’ll get when $k=16$. The coefficient won’t change when we multiply by the simple power $x^{-16}$, so the coefficient of $x^{16}$ in $(2x^4+5)^8$ will become the coefficient of $x^0$ in $x^{-16}(2x^4+5)^8$: the exponent will be reduced by $16$.
Knowing this, we can apply the binomial theorem to $(2x^4+5)^8$: it’s equal to
$$\sum_{i=0}^8\binom8{8-i}(2x^4)^{8-i}5^i=\sum_{i=0}^8\binom8{8-i}2^{8-i}x^{32-4i}5^i\;.$$
The exponent on $x$ is $16$ when $i=4$, and the coefficient is then
$$\binom8{8-4}2^{8-4}5^4=\binom842^45^4=10000\binom84=700,000\;.$$
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