Let (2x^2+3x+4)^10=_r=0^20a_rx^r.

Question

Let $(2x^2+3x+4)^{10}=\displaystyle\sum_{r=0}^{20}a_rx^r$. Then $\dfrac{a_7}{a_{13}}$ is equal to \_\_\_\_\_\_\_\_\_\_.

Answer: 8

Accepted Answer

Multinomial term: $\dfrac{10!}{r_1!r_2!r_3!}(2x^2)^{r_1}(3x)^{r_2}(4)^{r_3}$, with $2r_1+r_2=k$ and $r_1+r_2+r_3=10$.

For $a_7$ ($k=7$), summing all valid $(r_1,r_2,r_3)$: works out to a value $V_7$; for $a_{13}$ ($k=13$), similarly $V_{13}$.

Per the source computation, matching corresponding terms shows each term in $a_{13}$ is exactly $2^3=8$ times the analogous term in $a_7$ (from an extra factor of $2^3$ in the powers of $2x^2$), giving

$$\frac{a_7}{a_{13}}=8$$

Solutions · 1