If (1 - x + x^2 )^n = a_0 + a_1 x + a_2 x^2 + .... + a_ 2n x^ 2n … - Vidyadip

MCQ

If ${(1 - x + {x^2})^n} = {a_0} + {a_1}x + {a_2}{x^2} + .... + {a_{2n}}{x^{2n}}$, then ${a_0} + {a_2} + {a_4} + .... + {a_{2n}} = $

✓
$\frac{{{3^n} + 1}}{2}$
B
$\frac{{{3^n} - 1}}{2}$
C
$\frac{{1 - {3^n}}}{2}$
D
${3^n} + \frac{1}{2}$

✓

Answer

Correct option: A.

$\frac{{{3^n} + 1}}{2}$

a
(a) ${(1 - x + {x^2})^n} = {a_0} + {a_1}x + {a_2}{x^2} + .... + {a_{2n}}{x^{2n}}$

Putting $x = 1$, we get

${(1 - 1 + 1)^n} = {a_0} + {a_1} + {a_2} + ..... + {a_{2n}}$

==> $1 = {a_0} + {a_1} + {a_2} + .... + {a_{2n}}$.....$(i)$

Putting $x = -1,$ we get

==> ${3^n} = {a_0} - {a_1} + {a_2} - .... + {a_{2n}}$......$(ii)$

Adding $(i)$ and $(ii)$, we get

$\frac{{{3^n} + 1}}{2} = {a_0} + {a_2} + {a_4} + .... + {a_{2n}}$.

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