You don't have to find the aa's

Probability Level 5

xn=k=09aknk\large x_n=\sum_{k=0}^{9}a_k n^k

For the given values of the numbers aka_k for k=0,1,2,,9,k=0,1,2,\cdots, 9, it is true that xn=enx_n=e^n for n=1,2,,10.n=1, 2, \cdots, 10. Find x11.\left\lfloor x_{11}\right\rfloor.

Clarification: e=limn(1+1n)n2.71828 \displaystyle e = \lim_{n\to \infty} \left(1 +\dfrac1n \right)^n \approx 2.71828 .

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