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Question: Let {$a_n$} be a sequence defined by $a_1 = \frac{1}{2}$, $a_{n+1} = \frac{a_n^2}{a_n^2 - a_n + 1}$,...

Let {ana_n} be a sequence defined by a1=12a_1 = \frac{1}{2}, an+1=an2an2−an+1a_{n+1} = \frac{a_n^2}{a_n^2 - a_n + 1}, n≥1n \geq 1.

Find the exact value of S100=∑n=1100anS_{100} = \sum_{n=1}^{100} a_n.

A

1

B

100

C

1−1b101−11 - \frac{1}{b_{101}-1} where bnb_n is defined by b1=2,bn+1=bn2−bn+1b_1=2, b_{n+1}=b_n^2-b_n+1

D

0

Answer

1−1b101−11 - \frac{1}{b_{101}-1} where bnb_n is defined by b1=2,bn+1=bn2−bn+1b_1=2, b_{n+1}=b_n^2-b_n+1

Explanation

Solution

Let bn=1anb_n = \frac{1}{a_n}. Then b1=2b_1 = 2. The recurrence relation becomes: 1bn+1=(1/bn)2(1/bn)2−(1/bn)+1=1/bn21/bn2−1/bn+1=11−bn+bn2\frac{1}{b_{n+1}} = \frac{(1/b_n)^2}{(1/b_n)^2 - (1/b_n) + 1} = \frac{1/b_n^2}{1/b_n^2 - 1/b_n + 1} = \frac{1}{1 - b_n + b_n^2}. Thus, bn+1=bn2−bn+1b_{n+1} = b_n^2 - b_n + 1. Rearranging this, we get bn+1−1=bn2−bn=bn(bn−1)b_{n+1} - 1 = b_n^2 - b_n = b_n(b_n - 1). Now consider the term an=1bna_n = \frac{1}{b_n}. We can write: 1bn=bn−1bn(bn−1)=bn−1bn+1−1\frac{1}{b_n} = \frac{b_n - 1}{b_n(b_n - 1)} = \frac{b_n - 1}{b_{n+1} - 1}. This doesn't seem to lead to a simple form for ana_n.

Let's try another approach using the relation bn+1−1=bn(bn−1)b_{n+1} - 1 = b_n(b_n - 1). Consider the expression 1bn−1−1bn+1−1\frac{1}{b_n - 1} - \frac{1}{b_{n+1} - 1}. 1bn−1−1bn+1−1=1bn−1−1bn(bn−1)=bn−1bn(bn−1)=1bn\frac{1}{b_n - 1} - \frac{1}{b_{n+1} - 1} = \frac{1}{b_n - 1} - \frac{1}{b_n(b_n - 1)} = \frac{b_n - 1}{b_n(b_n - 1)} = \frac{1}{b_n}. So, an=1bn=1bn−1−1bn+1−1a_n = \frac{1}{b_n} = \frac{1}{b_n - 1} - \frac{1}{b_{n+1} - 1}.

Now we can compute the sum S100S_{100} as a telescoping series: S100=∑n=1100an=∑n=1100(1bn−1−1bn+1−1)S_{100} = \sum_{n=1}^{100} a_n = \sum_{n=1}^{100} \left( \frac{1}{b_n - 1} - \frac{1}{b_{n+1} - 1} \right) S100=(1b1−1−1b2−1)+(1b2−1−1b3−1)+⋯+(1b100−1−1b101−1)S_{100} = \left( \frac{1}{b_1 - 1} - \frac{1}{b_2 - 1} \right) + \left( \frac{1}{b_2 - 1} - \frac{1}{b_3 - 1} \right) + \dots + \left( \frac{1}{b_{100} - 1} - \frac{1}{b_{101} - 1} \right) The intermediate terms cancel out, leaving: S100=1b1−1−1b101−1S_{100} = \frac{1}{b_1 - 1} - \frac{1}{b_{101} - 1}. Since b1=2b_1 = 2, we have b1−1=2−1=1b_1 - 1 = 2 - 1 = 1. Therefore, S100=11−1b101−1=1−1b101−1S_{100} = \frac{1}{1} - \frac{1}{b_{101} - 1} = 1 - \frac{1}{b_{101} - 1}.