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7.Binomial Theorem
normal
If $(1 -x + x^2)^n = a_0 + a_1x + a_2x^2 + ....... + a_{2n}x^{2n}$, then $a_0 + a_2 + a_4 +........+ a_{2n}$ is equal to
A
$\frac{1}{2} (3^n+1)$
B
$\frac{1}{2} (3^n-1)$
C
$\frac{1}{2} (1-3^n)$
D
$\frac{1}{2} +3^n$
Solution
$\left(1-x+x^{2}\right)^{n}=a_{0}+a_{1} x+a_{2} x^{2}+\ldots \ldots \ldots a_{2 n} x^{2 n}$ ……..$(1)$
put $x=1,-1$ in $e{q^n}.{\rm{ }}\left( 1 \right)$ and add.
$a_{0}+a_{2}+\dots \dots \dots=\frac{3^{n}+1}{2}$
Standard 11
Mathematics
