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WBJEE · Maths · Binomial Theorem

Let \(\left(1+x+x^{2}\right)^{9}=a_{0}+a_{1} x+a_{2} x^{2}\)
\(+\ldots+a_{18} x^{18} \cdot\) Then

  1. A \(a_{0}+a_{2}+\ldots+a_{18}=a_{1}+a_{3}+\ldots+a_{17}\)
  2. B \(a_{0}+a_{2}+\ldots+a_{18}\) is even
  3. C \(a_{0}+a_{2}+\ldots+a_{18}\) is divisible by 9
  4. D \(a_{0}+a_{2}+\ldots+a_{18}\) is divisible by 3 but not by 9
Verified Solution

Answer & Solution

Correct Answer

(B) \(a_{0}+a_{2}+\ldots+a_{18}\) is even

Step-by-step Solution

Detailed explanation

\(\left(1+x+x^{2}\right)^{9}=a_{0}+a_{1} x+a_{2} x^{2}+ \dots+a_{18}x^{18}\) \[ Put \(x=-1,\) we get \] \begin{array}{l} (1-1+1)^{9}=a_{0}-a_{1}+a_{2}+\ldots+a_{18} \\ \Rightarrow 1=a_{0}-a_{1}+a_{2}+\ldots+a_{18} \end{array} \[ Put \(x=1,\) we get \]…