Let ( 2x^2 + 3x + 4 )^10 = _r = 0^20 a_rx^r Then a_7 a_13 is equal to ______.

$Note: Multinomial Theorem \\[1em] The general term of (x_1 + x_2 + + x_n)^n in the expansion is: \\ n!n_1! n_2! n_n! x_1^n_1 x_2^n_2 x_n^n_n \\[1em] where n_1 + n_2 + + n_n = n \\[1em] Here, in (2x^2 + 3x + 4)^10 the general term is: \\ aligned &= 10!n_1! n_2! n_3! (2x^2)^n_1 (3x)^n_2 (4)^n_3 \\ &= 10!n_1! n_2! n_3! 2^n_1 3^n_2 4^n_3 x^2n_1 + n_2

Let 2x2 3x 4 10 sumr 020 arxr Then a7 a13 is equal to : null (Mathematics)

\textbf{Note: Multinomial Theorem} \\[1em] \text{The general term of } (x_1 + x_2 + \dots + x_n)^n \text{ in the expansion is:} \\ \displaystyle \frac{n!}{n_1! n_2! \dots n_n!} x_1^{n_1} x_2^{n_2} \dots x_n^{n_n} \\[1em] \text{where } n_1 + n_2 + \dots + n_n = n \\[1em] \text{Here, in } (2x^2 + 3x + 4)^{10} \text{ the general term is:} \\ \begin{aligned} &= \frac{10!}{n_1! n_2! n_3!} (2x^2)^{n_1} (3x)^{n_2} (4)^{n_3} \\ &= \frac{10!}{n_1! n_2! n_3!} \cdot 2^{n_1} \cdot 3^{n_2} \cdot 4^{n_3} \cdot x^{2n_1 + n_2} \end{aligned} \\[1em] \therefore \text{ Coefficient of } x^{2n_1 + n_2} \text{ is:} \\ \displaystyle \frac{10!}{n_1! n_2! n_3!} \cdot 2^{n_1} \cdot 3^{n_2} \cdot 4^{n_3} \\[1em] \text{where } n_1 + n_2 + n_3 = 10 \\[1em] \text{For the coefficient of } x^7: \\ 2n_1 + n_2 = 7 \\[1em] \text{Possible values of } n_1, n_2, \text{ and } n_3 \text{ are:} \\ \begin{array}{|c|c|c|} \hline n_1 & n_2 & n_3 \\ \hline 3 & 1 & 6 \\ 2 & 3 & 5 \\ 1 & 5 & 4 \\ 0 & 7 & 3 \\ \hline \end{array} \\[1em] \therefore \text{ Coefficient of } x^7 \\ \begin{aligned} = \;& \frac{10!}{3!1!6!} (2)^3 (3)^1 (4)^6 + \frac{10!}{2!3!5!} (2)^2 (3)^3 (4)^5 \\ + \;& \frac{10!}{1!5!4!} (2)^1 (3)^5 (4)^4 + \frac{10!}{0!7!3!} (2)^0 (3)^7 (4)^3 \end{aligned} \\[1em] \text{Coefficient of } x^{13} = a_{13} \\ \text{Here } 2n_1 + n_2 = 13 \\[1em] \text{Possible values of } n_1, n_2, \text{ and } n_3 \text{ are:} \\ \begin{array}{|c|c|c|} \hline n_1 & n_2 & n_3 \\ \hline 6 & 1 & 3 \\ 5 & 3 & 2 \\ 4 & 5 & 1 \\ 3 & 7 & 0 \\ \hline \end{array} \\[1em] \therefore \text{ Coefficient of } x^{13} \\ \begin{aligned} = \;& \frac{10!}{6!1!3!} (2)^6 (3)^1 (4)^3 + \frac{10!}{5!3!2!} (2)^5 (3)^3 (4)^2 \\ + \;& \frac{10!}{4!5!1!} (2)^4 (3)^5 (4)^1 + \frac{10!}{3!7!0!} (2)^3 (3)^7 (4)^0 \end{aligned} \\[1em] \therefore \frac{a_7}{a_{13}} = 8

Explanation

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