For all integers $n \geq 1$, define $a(n)=\left[\log _n(2002)\right]^{-1}$. Let $b=a(2)+a(3)+a(4)+a(

Question

For all integers $n \geq 1$, define $a(n)=\left[\log _n(2002)\right]^{-1}$. Let $b=a(2)+a(3)+a(4)+a(5)$ and $c=a(10)+a(11)+a(12)+a(13)+a(14)$. Then $b-c$ is equal to

Professor PC Thomas Classes · Answer

$a(n) = \log_{2002}(n)$. Combining yields $b = \log_{2002}(120)$ and $c = \log_{2002}(240240)$. Subtracting equates to $\log_{2002}(\frac{1}{2002}) = -1$.