If {(1 - x + {x^2})^n} = {a_0} + {a₁}x + {a₂}{x^2} + .... + {a_{2n}}{x^{2n}} , then {a_0} + {a?

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7.Binomial Theorem

hard

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}} = $

A

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

B

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

C

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

D

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

Solution

(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

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

Standard 11

Mathematics

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