If a_ n = ( _ e 3 )^ n _ k=0 ^ n 1 k!(n-k)! , for n ≥ 0 then a_ n +a_ 1 +a_ 2 + =
Mathematics · JEE Main · NTA Exams — Binomial Theorem And Its Simple Applications
If \(a_{n}=\left(\log _{e} 3\right)^{n} \sum_{k=0}^{n} \frac{1}{k!(n-k)!}\), for n ≥ 0 then \(a_{n}+a_{1}+a_{2}+\ldots \infty\) =
- 2n
- 9
- \(\frac{z^{n}}{\{n-1\}!}\)
- None of these
Answer
(D) None of these
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