Gujarati
7.Binomial Theorem
hard

यदि ${(1 - x + {x^2})^n} = {a_0} + {a_1}x + {a_2}{x^2} + .... + {a_{2n}}{x^{2n}}$, तो ${a_0} + {a_2} + {a_4} + .... + {a_{2n}}$ बराबर है

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

$x = 1$ रखने पर,

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

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

$x = -1$ रखने पर,

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

$(i)$ व $(ii)$ जोड़ने पर,

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

Std 11
Mathematics

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