MCQ
If $\left(1-x+x^2\right)^n=a_0+a_1 x+a_2 x^2+\ldots+a_{2 n} x^{2 n}$, then $a_0+a_2+a_4+\ldots+a_{2 n}$ equals.
  • A
    $3^n+\frac{1}{2}$
  • B
    $\frac{3^n+1}{2}$
  • C
    $\frac{3^n-1}{2}$
  • D
    $\frac{1-3^n}{2}$

Answer

(b) $\frac{x^2+1}{2}$
Explanation: $\left(1-x+x^2\right)^n=a_0+a_1 x+a_2 x^2+\ldots+a_{2 n} x^{2 n} . .(1)$
Put x=1 in (1),we get
$1=a_0+a_1+a_2+a_3+\ldots+a_{2 n} . .(2)$
Put x=-1 in(1),we get
3^n=a_0-a_1+a_2-a_3+\ldots+a_{2 n}
Adding(1) and(2),we get
$3^n+1=2\left(a_0+a_2+a_4+\ldots+a_{2 n}\right)$
Thus, $a_0+a_2+a_4+\ldots+a_{2 n}=\frac{3^n+1}{2}$

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