A question on the multinomial expansion

Question

Given that $$(2x^2+3x+4)^{10}=\sum_{i=0}^{20} a_{i}x^{i} $$ Calculate the value of $\frac {a_{7}}{a_{13}} $.

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I have manually taken all the cases of formation of $x^7$ and $x^{13}$ and arrived at the answer 8. But definitely there must be a pattern in the ratio of $\frac {a_i}{a_{20-i}}$ which I am not able to find. My guess is (just a guess) that $\frac {a_i}{a_{20-i}} = 2^{10-i} $ .

Answer 1

Let $\displaystyle f(x)=(2x^2+3x+4)^{10}=\sum_{i=0}^{20} a_{i}x^{i}$.

Then, $\displaystyle f(2x)=(8x^2+6x+4)^{10}=2^{10}(4x^2+3x+2)^{10}=2^{10}x^{20}f\Big(\frac{1}{x}\Big)$.

This implies that $2^ia_i=2^{10}a_{20-i}$, which confirms your guess.

Answer 2

With $f(x):=(2x^2+3x+4)^{10}$, we observe that (for $x\ne0$) $$x^{20} f(\tfrac 2x)=x^{20}(8x^{-2}+6x^{-1}+4)^{10}={2^{10}}(4+3x+2x^2)^{10}={2^{10}}f(x).$$ Expanding both sides, $$ x^{20}\sum_{k=0}^{20}a_k2^kx^{-k}=2^{10}\sum_{k=0}^{20}a_kx^k$$ $$ \sum_{k=0}^{20}a_k2^kx^{20-k}=\sum_{k=0}^{20}2^{10}a_kx^k$$ $$ \sum_{k=0}^{20}a_{20-k}2^{20-k}x^{k}=\sum_{k=0}^{20}2^{10}a_kx^k$$ $$ a_{20-k}2^{20-k}=2^{10}a_k$$ and finally $$ \frac{a_{k}}{a_{20-k}}=2^{10-k}$$ (or $a_k=a_{20-k}=0$).