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EAMCET Engineering 2010 Solved Paper

Section: Mathematics

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Question : 57 of 160

Marks: +1, -0

\;\text{ If }\;(1+2 x+3 x^2)^{10}

\;\; =a_0+a_1 x+a_2 x^2+\ldots +a_{20} x^{20}

\;\frac{a_2}{a_1}

10.5
21
10
5.5

Solution:

\;(1+2 x+3 x^2)^{10}

\;\; =a_0+a_1 x+a_2 x^2+\ldots +a_{20} x^{20}

=\{1+x(2+3x)\}^{10}

Then by binomial expansion

={}^{10}C_0+{}^{10}C_1 x(2+3x)+{}^{10}C_2 x^2(2+3x)^2+\ldots

Now, the coefficient of

x

in this expansion ie,

\;\; a_1=2 \cdot 10=20

and the coefficient of

x^2

in this expansion

=3 \cdot {}^{10}C_1+4 \cdot {}^{10}C_2

i e

a_2=3 \cdot 10+4 \cdot 45

=30+180

=210

So,

\;\;\; \frac{a_2}{a_1}=\;\frac{210}{20}=10.5

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