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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,
$\frac{{{3^n} + 1}}{2} = {a_0} + {a_2} + {a_4} + …. + {a_{2n}}$.
Standard 11
Mathematics
