If (1+2x+3x^2)^{10} = a_0+a_1x+a_2x^2+ldots+a | vedclass

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(A) Given $(1+2x+3x^2)^{10} = a_0+a_1x+a_2x^2+\ldots+a_{20}x^{20}$.Using the binomial expansion $(1+y)^n = \sum_{k=0}^{n} {}^{n}C_k y^k$,where $y = 2x

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$10.5$

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(A) Given $(1+2x+3x^2)^{10} = a_0+a_1x+a_2x^2+\ldots+a_{20}x^{20}$.Using the binomial expansion $(1+y)^n = \sum_{k=0}^{n} {}^{n}C_k y^k$,where $y = 2x+3x^2$:$(1+(2x+3x^2))^{10} = {}^{10}C_0 + {}^{10}C_1(2x+3x^2) + {}^{10}C_2(2x+3x^2)^2 + \ldots$$= 1 + 10(2x+3x^2) + 45(4x^2+12x^3+9x^4) + \ldots$$= 1 + 20x + 30x^2 + 180x^2 + 540x^3 + 405x^4 + \ldots$$= 1 + 20x + 210x^2 + \ldots$Comparing coefficients:$a_1 = 20$$a_2 = 210$Therefore,$\frac{a_2}{a_1} = \frac{210}{20} = 10.5$.

If $(1+2x+3x^2)^{10} = a_0+a_1x+a_2x^2+\ldots+a_{20}x^{20}$,then $\frac{a_2}{a_1}$ is equal to

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