Question

A motorcyclist starts at noon and drives 10 miles in the first hour, and then in each hour after that, he drives 6 more miles than in the previous hour. After how many hours of driving will the motorcyclist drive a total of 248 miles?In what hour of driving will the motorcyclist cross the 400 mile mark?

Answer 1

Notice the following pattern:

`\begin{gathered} \text{first hour=10} \\ \text{ second hour=10+6=16} \\ \text{third hour=(10+6)+6=22} \end{gathered}`

Then, the function that gives us the miles the motorcyclist drives in a time t (t in hours) is

`\begin{gathered} d(t)=10+(t-1)6,t\ge1 \\ \Rightarrow d(t)=4+6t,t\ge1 \end{gathered}`

a) We need to find t such that

`\begin{gathered} d(t)+d(t-1)+\cdots+d(1)=248 \\ \end{gathered}`

Notice that:

`\begin{gathered} Pattern\text{ of driven miles} \\ 10 \\ 10+16 \\ 10+16+22 \\ 10+16+22+28 \\ 10+16+22+28+34=110 \\ 10+16+22+28+34+40=150 \\ 10+16+22+28+34+40+46=196 \\ 10+16+22+28+34+40+46+52=248 \end{gathered}`

Then, the answer is 8 hours.

b)

We can continue the pattern given above and obtain

`\begin{gathered} 10+16+22+28+34+40+46+52+58=306 \\ 10+16+22+28+34+40+46+52+58+64=370 \\ 10+16+22+28+34+40+46+52+58+64+70=440 \end{gathered}`

Then, the answer is the 11th hour

Solving the problem using summation notation:

`\text{traveled distance after n hours}=\sum ^n_(t=1)(4+6t)`

Therefore

`\sum ^n_(t=1)(4+6t)=\sum ^n_(t=1)4+\sum ^n_(t=1)6t=4n+6\sum ^n_(t=1)t=4n+6((n(n+1))/(2))=4n+3n(n+1)`

Thus,

`Traveled.dis\tan ce.after.n.hours=4n+3n(n+1)`

In the last step, we used the Gauss sums of consecutive integers., which states that

`\sum ^n_(x=1)x=(n(n+1))/(2)`

The formula to know how many miles we have traveled after n hours is 4n+3n(n+1)