1+2x+3x210=a0+a1x+a2x2+…+a20x20⇒a2a1=

a0+a1x+a2x2+…+a20x20=1+2x+3x210=10C0+10C12x+3x2+10C22x+3x22+…=1+102x+3x2+454x2+12x3+9x4+…=1+20x+210x2+…⇒a0=1,a1=20,a2=210,…⇒a2a1=21020=10⋅5

1+2x+3x210=a0+a1x+a2x2+…+a20x20⇒a2a1=

a0+a1x+a2x2+…+a20x20=1+2x+3x210=10C0+10C12x+3x2+10C22x+3x22+…=1+102x+3x2+454x2+12x3+9x4+…=1+20x+210x2+…⇒a0=1,a1=20,a2=210,…⇒a2a1=21020=10⋅5

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${(1 + 2 x + 3 x^{2})}^{10} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{20} x^{20} \Rightarrow \frac{a_{2}}{a_{1}} =$

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Detailed Solution

$\begin{array}{l} a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{20} x^{20} = {(1 + 2 x + 3 x^{2})}^{10} =^{10} C_{0} +^{10} C_{1} (2 x + 3 x^{2}) +^{10} C_{2} {(2 x + 3 x^{2})}^{2} + \dots \\ = 1 + 10 (2 x + 3 x^{2}) + 45 (4 x^{2} + 12 x^{3} + 9 x^{4}) + \dots = 1 + 20 x + 210 x^{2} + \dots \\ \Rightarrow a_{0} = 1, a_{1} = 20, a_{2} = 210, \dots \Rightarrow \frac{a_{2}}{a_{1}} = \frac{210}{20} = 10 \cdot 5 \end{array}$