WBJEE · Maths · Binomial Theorem
If \(\left(1-x+x^2\right)^n=a_0+a_1 x+\ldots . .+a_{2 n} x^{2 n}\), then the value of \(a_0+a_2+a_4+\ldots \ldots .+a_{2 n}\) is
- A \(3^{\mathrm{n}}+\frac{1}{2}\)
- B \(3^{\mathrm{n}}-\frac{1}{2}\)
- C \(\frac{3^{\mathrm{n}}-1}{2}\)
- D \(\frac{3^{\mathrm{n}}+1}{2}\)
Answer & Solution
Correct Answer
(D) \(\frac{3^{\mathrm{n}}+1}{2}\)
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