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Q. If $\left(1-x+x^{2}\right)^{n} =a_{0} +a_{1}x +...+\,a_{2n}x^{2n} , $ then $a_{0} +\,a_{2} +a_{4}\,+....+\,a_{2n}=$

COMEDKCOMEDK 2011Binomial Theorem

Solution:

$\left(1-x+x^{2}\right)^{n}$ $ =a_{0} +a_{1}x +a_2x^2 + ...+ \,a_{2n}x^{2n}$ ...(i)
Putting x = 1 in (i), we get $1 = a_{0} +a_{1} +a_{2}+ .... +a_{2n}$ ....(ii)
Putting x = -1 in (i), we get $3^n = a_{0} +a_{1} +a_{2} - a_{3} +.... +a_{2n}$ ....(iii)
Adding (ii) and (iii), we get $1+3^{n} =2a_{0 }+2a_{2} +2a_{4} +...+2a_{2n}$
$ \Rightarrow a_{0}+a_{2}+a_{4}+....a_{2n} =\left(\frac{3^{n}+1}{2}\right)$