Solve the following


If (1 – x + x2)n = a0 + a1 x + a2 x2 +...+a2n  x2n, then a0 + a2 + a4 +...+ a2n equals

(a) $\frac{3^{n}+1}{2}$

(b) $\frac{3^{n}-1}{2}$

(c) $\frac{1-3^{n}}{2}$

(d) $3^{n}+\frac{1}{2}$




$\left(1-x+x^{2}\right) n=a_{0}+a_{1} x+a_{2} x^{2}+\ldots+a_{2 n} x^{2 n} \quad \ldots(1)$

In order to find $a_{0}+a_{2}+a_{4}+\ldots+a_{2 n}$

i.e. all even terms are involved

∴ replace x by 1 in equation (1)

we get

$(1-1+1)^{n}=a_{0}+a_{1}+a_{2}+\ldots \ldots+a_{2 n}$

i.e. $1=a_{0}+a_{1}+a_{2}+\ldots+a_{2 n} \quad \ldots$ (2)

and now replace by –1 in equation (1), we get

$(1+1+1)^{n}=a_{0}-a_{1}+a_{2}-a_{3}+\ldots+a_{2 n}$

i.e. $3^{n}=a_{0}-a_{1}+a_{2}-a_{3}+\ldots \ldots+a_{2 n} \quad \ldots(3)$

By adding (2) and (3), we get

$3^{n}+1=2 a_{0}+2 a_{2}+2 a_{4}+\ldots+2 a_{2 n}$

i.e. $a_{0}+a_{2}+a_{4}+\ldots+\mathrm{a}_{2 n}=\frac{1+3^{n}}{2}$

Hence, the correct answer is option A.



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