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Alecsey [184]
3 years ago
13

Can someone help me with this? **SHOW WORK PLEASE**

Mathematics
1 answer:
Valentin [98]3 years ago
3 0

FYI, that L in the denominator is factorial.

This is some pretty serious stuff for high school.

My inclination is that the answer is A, which is the definition of <em>e. </em> Let's see if we can show this.

Let's write

\displaystyle f(n) = \left(1 + \dfrac 1 n \right)^n

Let's expand this with the binomial expansion

\displaystyle f(n) =\sum_{k=0}^n {n \choose k} \dfrac{1}{n^k}

\displaystyle f(n) =\sum_{k=0}^n \dfrac{n!}{k!(n-k)!} \cdot \dfrac{1}{n^k}

\displaystyle f(n) =\sum_{k=0}^n \dfrac{n(n-1)\cdots(n-k+1)}{k! \, n^k}

Let's focus on when n is really big and on the ks that are relatively small, which make up the bulk of the sum as the terms get small rapidly.

Then that numerator n(n-1)···(n-k+1) ≈ n^k as all the factors are about n.  

\displaystyle f(n)\approx \sum_{k=0}^n \dfrac{n^k}{k!\, n^k}

\displaystyle f(n)\approx \sum_{k=0}^n \dfrac{1}{k!}

OK, we showed for large n this is approximately true, and it will be exactly true in the limit, so we choose

Answer: A

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