AP EAMCET · Maths · Binomial Theorem
If \(\left(1+x+x^2\right)^n=a_0+a_1 x+a_2 x^2+\ldots+a_{2 n} x^{2 n}\), then \(a_0+a_2+a_4+\ldots+a_{2 n}=\)
- A \(3^n\)
- B \(3^n+1\)
- C \(\frac{3^n-1}{2}\)
- D \(\frac{3^n+1}{2}\)
Answer & Solution
Correct Answer
(D) \(\frac{3^n+1}{2}\)
Step-by-step Solution
Detailed explanation
\(P(x)=\left(1+x+x^2\right)^n=a_0+a_1 x+a_2 x^2+\ldots+a_{2 n} x^{2 n}\) \(a_0+a_2+a_4+\ldots+a_{2 n}=\frac{P(1)+P(-1)}{2}\) \(P(1)=(1+1+1^2)^n=3^n\) \(P(-1)=(1-1+(-1)^2)^n=1^n=1\) \(a_0+a_2+a_4+\ldots+a_{2 n}=\frac{3^n+1}{2}\)
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