Problem Solving

Manhattan Prep

$$\text{For all positive integers n, the sequence } A_n \text{ is defined by the following relationship:}$$
$$A_n=\frac{n-1}{n!}$$
$$\text{What is the sum of all the terms in the sequence from } A_1 \text{ through } A_{10}, \text{ inclusive?}$$
$$\text{A. }\frac{9!+1}{10!}$$
$$\text{B. }\frac{9(9!)}{10!}$$
$$\text{C. }\frac{10!-1}{10!}$$
$$\text{D. }\frac{10!}{10!+1}$$
$$\text{E. }\frac{10(10!)}{11!}$$
OA C

AAPL wrote:Manhattan Prep

$$\text{For all positive integers n, the sequence } A_n \text{ is defined by the following relationship:}$$
$$A_n=\frac{n-1}{n!}$$
$$\text{What is the sum of all the terms in the sequence from } A_1 \text{ through } A_{10}, \text{ inclusive?}$$
$$\text{A. }\frac{9!+1}{10!}$$
$$\text{B. }\frac{9(9!)}{10!}$$
$$\text{C. }\frac{10!-1}{10!}$$
$$\text{D. }\frac{10!}{10!+1}$$
$$\text{E. }\frac{10(10!)}{11!}$$
OA C

Try developing a pattern.

First number in sequence is (1-1)/1 = 0
second is 1/2
third is (3-1)/3! = 1/3

The sum of this series is 1/2 + 1/3 = 5/6

Test some answers recognizing that since 3 numbers in the sequence, 2 corresponds to 9, 3 corresponds to 10 and 4 to 11

Try A: ( 2!+ 1)/3! = 3/6 =1/2 > no match
B: 2(2!)/3! = 4/6=2/3 > no match
C: (3!-1)/3! = (6-1)/6 = 5/6 > We have a winner