Determine the middle term in the expansion of $(2x+3^y)$ to the power of $12$.
I've tried to do it but I only know how without the $2$ that's in front .
2026-04-03 21:25:32.1775251532
The middle term of $(2x+3^y)^{12}$
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HINT: The binomial theorem says that $$(a+b)^{12}=\sum_{k=0}^{12}\binom{12}ka^kb^{12-k}\;,$$ the middle term being the $k=6$ term, $$\binom{12}6a^6b^6\;.\tag{1}$$ You have $(2x+3^y)^{12}$, so in your case $a=2x$ and $b=3^y$; just substitute those into $(1)$ and simplify the result.