سوال ۱۱۲ کنکور ارشد علوم کامپیوتر ۱۴۰۵

The recurrence is multiplicative: each term is the previous one scaled by $b_n$.

$a_n = b_n\,a_{n-1}, \qquad n \ge 2.$

Unroll it (telescoping product):

$a_2 = b_2 a_1,\quad a_3 = b_3 a_2 = b_3 b_2 a_1,\ \dots,\ a_n = b_n b_{n-1}\cdots b_2\, a_1.$

Equivalently, multiply the consecutive ratios so everything cancels:

$\frac{a_n}{a_1} = \prod_{i=2}^{n}\frac{a_i}{a_{i-1}} = \prod_{i=2}^{n} b_i \;\Longrightarrow\; a_n = a_1\prod_{i=2}^{n} b_i.$

This is option 3.

• Options 1 and 2 are wrong because the recurrence is a product, not a sum. (An additive recurrence $a_n=a_{n-1}+b_n$ would give $a_1+\sum b_i$, not $a_1\sum b_i$.)
• Option 4 is wrong because $b_1$ never appears: the recurrence is defined only for $n \ge 2$, so the accumulating factors are $b_2,\dots,b_n$ and the product must start at $i=2$.

Exam trick: In a telescoping product the base term $a_1$ is already fixed, so the product index starts one step after the base ($i=2$). The lower limit is exactly what separates the correct option from the $i=1$ distractor.