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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Solution

The correct option is B

Put x = 1,

⇒ $(1 + 1 + 2)^{20} = a_{0} + a_{1} + a_{2} + a_{3} + a_{4} . . . . . . . . . . . . . . . a_{40}$ ----------------(1)

Put x = -1

⇒ $(1 - 1 + 2)^{20} = a_{0} + a_{1} + a_{2} + a_{3} + a_{4} . . . . . . . . . . . . . . . a_{40}$ ----------------(2)

(1) + (2) ⇒ $4^{20} + 2^{20}$ = $2 [a_{0} + a_{2} + a_{4} . . . . . . . . . . . . . . . . . a_{38} + a_{40}]$

⇒ $(a_{0} + a_{2} + a_{4} . . . . . . . . . . . . . . . a_{38} + a_{40})$ = $\frac{2^{40} + 2^{20}}{2}$ = $2^{39} + 2^{19}$

$a_{40}$ is the coefficient of $x^{40}$ . It will be $2^{20}$ , because the last term in the expansion will be $2^{20} (x^{2})^{20}$ .

⇒ $(a_{0} + a_{2} + a_{4} . . . . . . . . . . . . . . . a_{38} = 2^{39} + 2^{19} - 2^{20}$

= $2^{19} (2^{20} + 1 - 2)$

= $2^{19} (2^{20} - 1)$

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If (1+x+2x2)20=a0+a1x+a2x2.....a40x40, then a0+a2+a4....a38 =

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