For all integers n greater than 1, define an=logn20021. Let b=a2+a3+a4+a5 and c=a10+a11+a12+a13+a14. Find b−c.
Solution — click to reveal
We have an=logn20021=log2002n, so b−c==(log20022+log20023+log20024+log20025)−(log200210+log200211+log200212+log200213+log200214)log200210⋅11⋅12⋅13⋅142⋅3⋅4⋅5=log200211⋅13⋅141=log200220021=−1.