AP DSC School Assistant Math Model Test 2

Concept:The series is S = 1 + 2³/2! + 3³/3! + 4³/4! + … which equals $\sum_{n=1}^{\infty} \frac{n^3}{n!}$. We rewrite $n^3$ as $n(n-1)(n-2) + 3n(n-1) + n$ and use the standard exponential series $\sum_{n=0}^\infty \frac{x^n}{n!} = e^x$ with $x = 1$.Explanation:Step 1: Express $n^3$ in a form that cancels factorials:$$\frac{n^3}{n!} = \frac{n(n-1)(n-2) + 3n(n-1) + n}{n!}.$$Step 2: Split the sum into three separate series:$$S = \sum_{n=3}^{\infty} \frac{n(n-1)(n-2)}{n!} + 3\sum_{n=2}^{\infty} \frac{n(n-1)}{n!} + \sum_{n=1}^{\infty} \frac{n}{n!}.$$(The lower limits adjust because terms vanish for small $n$.)Step 3: Simplify each term using factorial cancellation:$\frac{n(n-1)(n-2)}{n!} = \frac{1}{(n-3)!}$ for $n \ge 3$,$\frac{n(n-1)}{n!} = \frac{1}{(n-2)!}$ for $n \ge 2$,$\frac{n}{n!} = \frac{1}{(n-1)!}$ for $n \ge 1$.Thus$$S = \sum_{n=3}^{\infty} \frac{1}{(n-3)!} + 3\sum_{n=2}^{\infty} \frac{1}{(n-2)!} + \sum_{n=1}^{\infty} \frac{1}{(n-1)!}.$$Step 4: Shift indices to start from $0$:First sum: let $k = n-3$, then $\sum_{k=0}^{\infty} \frac{1}{k!} = e$.Second sum: let $k = n-2$, then $3 \sum_{k=0}^{\infty} \frac{1}{k!} = 3e$.Third sum: let $k = n-1$, then $\sum_{k=0}^{\infty} \frac{1}{k!} = e$.Step 5: Add them:$$S = e + 3e + e = 5e.$$Step 6: Compare with the given options:A. $5e - 1$, B. $5e - 3$, C. $4e$. None equals $5e$. Therefore, the sum is not among A, B, or C.Answer:Option D: None of these.