The remainder after n terms is R_n=S_infinity-S_n=arn/(1-r)=S_infinity rn. Its magnitude is |S_infinity|,|r|n. Subtracting S_n=S_infinity(1-rn) from the limiting total leaves exactly S_infinity rn. For a negative ratio, the signed remainder changes sign. Find the limiting total, form the remainder or its magnitude, solve the accuracy condition, and check neighboring integer counts. In a journey, separate an initial one-way segment from repeated up-and-down segments. For a=9,r=-1/2, S_infinity=6 and R_3=6(-1/2)3=-3/4. Thus S_3=27/4 lies above its limit. Exam check: a ball dropped from height h, rebounding to fraction r of each previous height, travels total distance h+2hr/(1-r), not 2h/(1-r).

The remainder after \(n\) terms is \(R_n=S_\infty-S_n=\dfrac{ar^n}{1-r}=S_\infty r^n\). Its magnitude is \(|S_\infty|\,|r|^n\).

Subtracting \(S_n=S_\infty(1-r^n)\) from the limiting total leaves exactly \(S_\infty r^n\). For a negative ratio, the signed remainder changes sign.

Find the limiting total, form the remainder or its magnitude, solve the accuracy condition, and check neighboring integer counts. In a journey, separate an initial one-way segment from repeated up-and-down segments.

For \(a=9,r=-\dfrac{1}{2}\), \(S_\infty=6\) and \(R_3=6\left(-\dfrac{1}{2}\right)^3=-\dfrac{3}{4}\). Thus \(S_3=\dfrac{27}{4}\) lies above its limit.

Exam check: a ball dropped from height \(h\), rebounding to fraction \(r\) of each previous height, travels total distance \(h+\dfrac{2hr}{1-r}\), not \(\dfrac{2h}{1-r}\).

Remainders and repeated journeys

Remainders and repeated journeys