\[\large x_n=\sum_{k=0}^{9}a_k n^k\]

For the given values of the numbers \(a_k\) for \(k=0,1,2,\cdots, 9,\) it is true that \(x_n=e^n\) for \(n=1, 2, \cdots, 10.\) Find \(\left\lfloor x_{11}\right\rfloor.\)

**Clarification**:
\( \displaystyle e = \lim_{n\to \infty} \left(1 +\dfrac1n \right)^n \approx 2.71828 \).

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