Let $(1 + x + x^{2} + x^{3} + x^{4})^{10} = a_{0} + a_{1} x + a_{2} x^{2} + . . . + a_{40} x^{40}$ . Then:

a_{1} = 20

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a_{2} = 110

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a_{2} = 55

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a_{1} + a_{2} = 65

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Solution

The correct options are
C $a_{2} = 55$
D $a_{1} + a_{2} = 65$
Coefficient of $x^{1}$ will be
$= \frac{10!}{1! .9!} = 10$ .
Now $x^{2}$ can be achieved in the following way
$1^{8} . x^{2} . (x^{2})^{0} . (x^{3})^{0} . (x^{4})^{0}$
Or
$1^{9} . x^{0} . (x^{2})^{1} . (x^{3})^{0} . (x^{4})^{0}$
Hence coefficients will be
$= \frac{10!}{2! .8!} + \frac{10!}{1! .9!}$
$=^{10} C_{8} +^{10} C_{9}$
$=^{11} C_{9}$
$= 55$ .
Hence
$a_{2} + a_{1} = 55 + 10 = 65$ .

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If $(1 + x + 2 x^{2})^{20} = a_{0} + a_{1} x + a_{2} x^{2} . . . . . a_{40} x^{40}$ , then $a_{0} + a_{2} + a_{4} . . . . a_{38}$ =

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Let (1+x+x2+x3+x4)10=a0+a1x+a2x2+...+a40x40. Then:

The correct options are C a2=55 D a1+a2=65Coefficient of x1 will be =10!1!.9!=10.Now x2 can be achieved in the following way18.x2.(x2)0.(x3)0.(x4)0Or 19.x0.(x2)1.(x3)0.(x4)0Hence coefficients will be =10!2!.8!+10!1!.9!=10C8+10C9=11C9=55.Hencea2+a1=55+10=65.

Let $(1 + x + x^{2} + x^{3} + x^{4})^{10} = a_{0} + a_{1} x + a_{2} x^{2} + . . . + a_{40} x^{40}$ . Then: