20
The sequence given is a series of products of consecutive odd numbers taken three at a time. The general term for the sequence can be expressed as \( (2k-1)(2k+1)(2k+3) \), where \( k \) is the term number starting from 1. To find \( f(2) \), consider the sum of the first two terms of the series: \[ 1\cdot3\cdot5 + 3\cdot5\cdot7 = 15 + 105 = 120 \] Given that this sum is expressed by the formula \( n(n+1)f(n) \) for \( n = 2 \), we have: \[ 2(2+1)f(2) = 120 \quad \Rightarrow \quad 6f(2) = 120 \quad \Rightarrow \quad f(2) = 20 \] Thus, \( f(2) = 20 \), which confirms the correct answer as option (4).
For \( n \in \mathbb{N} \), the largest positive integer that divides \( 81^n + 20n - 1 \) is \( k \). If \( S \) is the sum of all positive divisors of \( k \), then find \( S - k \).
Match the following: