Infinite Geometric Series: Additional Practice (Set 1)

Substituting $n = 1$ gives the first term $\dfrac{5}{36}$, and each step multiplies the numerator by $5$ and the denominator by $6$, so $r = \tfrac56$.

The structure is easier to see after pulling out a constant: since $6^{\,n+1} = 6 \cdot 6^{\,n}$, the general term is $\dfrac{1}{6}\left(\dfrac56\right)^{\,n}$.

$S = \frac{\tfrac{5}{36}}{1 - \tfrac56} = \frac{\tfrac{5}{36}}{\tfrac16} = \frac{5}{36}\cdot 6 = \frac56$

A ratio of $\tfrac56$ is close to $1$, so this series converges slowly, but it converges.

Track what is left of each past dose at the moment of the newest one. Today's dose contributes $200$ mg, yesterday's has been cut to $200\left(\tfrac14\right)$, the one before to $200\left(\tfrac14\right)^2$, and so on.

$200 + 200\left(\tfrac14\right) + 200\left(\tfrac14\right)^2 + \cdots$

That is geometric with $a = 200$ and $r = \tfrac14$, and $\left|\tfrac14\right| < 1$, so it converges.

$S = \frac{200}{1 - \tfrac14} = \frac{200}{\tfrac34} = \frac{800}{3} \approx 266.7 \text{ mg}$

The level never exceeds this, and it never quite reaches it either. It closes in on it, which is exactly what a convergent series describes.

Separate the trip into the first drop and the rebounds, because the first drop happens only once while every rebound height is travelled twice, once up and once down.

The rebound heights are $6$, then $6\left(\tfrac35\right)$, and so on, a geometric series with $a = 6$ and $r = \tfrac35$.

$\text{sum of rebound heights} = \frac{6}{1 - \tfrac35} = \frac{6}{\tfrac25} = 15$

Now assemble the total distance. It is the initial $10$ meter drop plus twice the rebound total.

$10 + 2(15) = 40 \text{ meters}$

Forgetting the factor of $2$ gives $25$, and forgetting the initial drop gives $30$. Both are on the list.

The $4$ does not repeat, so peel it off and treat only the $5$s as a series.

$0.4\overline{5} = \frac{4}{10} + \left(\frac{5}{100} + \frac{5}{1000} + \cdots\right)$

The bracket has $a = \tfrac{5}{100}$ and $r = \tfrac{1}{10}$, so it sums to $\dfrac{5/100}{9/10} = \dfrac{5}{100}\cdot\dfrac{10}{9} = \dfrac{1}{18}$.

$0.4\overline{5} = \frac{2}{5} + \frac{1}{18} = \frac{36}{90} + \frac{5}{90} = \frac{41}{90}$

The decoy $\tfrac{45}{99}$ treats $45$ as the repeating block, but that is $0.454545\ldots$, and $\tfrac{9}{20}$ is the terminating $0.45$.

The first term is $\left(\tfrac{x+2}{5}\right)^0 = 1$, never zero, so the criterion applies to the ratio $r = \tfrac{x+2}{5}$.

$\left|\frac{x+2}{5}\right| < 1 \;\Longleftrightarrow\; |x+2| < 5$

Unpack the absolute value into a double inequality and subtract $2$ throughout.

$-5 < x + 2 < 5 \;\Longleftrightarrow\; -7 < x < 3$

The interval is centred at $x = -2$, where the ratio is $0$, and it reaches $5$ units in each direction. Sliding the centre is what the $+2$ does.

Evaluate the ratio at $x = 1$.

$r = \frac{1+2}{5} = \frac35$

That value lies inside $-7 < x < 3$, so the series converges, and its first term is $1$.

$S = \frac{1}{1 - \tfrac35} = \frac{1}{\tfrac25} = \frac52$

Dividing by $r$ rather than $1 - r$ would give $\tfrac53$, which is the trap here.

Halving every side halves the perimeter, so the perimeters are $36$, $18$, $9$, and so on: a geometric series with $a = 36$ and $r = \tfrac12$.

$S = \frac{36}{1 - \tfrac12} = \frac{36}{\tfrac12} = 72$

Any geometric series with ratio $\tfrac12$ totals exactly twice its first term, so the infinitely many triangles have the same total perimeter as two copies of the original.

The areas are $100$, then $25$, then $\tfrac{25}{4}$, and so on, so this is geometric with $a = 100$ and $r = \tfrac14$.

$S = \frac{100}{1 - \tfrac14} = \frac{100}{\tfrac34} = \frac{400}{3} \approx 133.3$

The answer $\tfrac{100}{3}$ is the total of the inscribed triangles only, that is, the sum with the first term left out, and it is worth checking that $100 + \tfrac{100}{3} = \tfrac{400}{3}$ as it should.

The index starts at $n = 2$, so the first term of this sum is the $n = 2$ term, and $a$ in the formula must be that term.

$a = 3\left(\tfrac25\right)^2 = 3 \cdot \frac{4}{25} = \frac{12}{25}, \qquad r = \frac25$

Now apply the formula.

$S = \frac{\tfrac{12}{25}}{1 - \tfrac25} = \frac{\tfrac{12}{25}}{\tfrac35} = \frac{12}{25}\cdot\frac53 = \frac45$

The value $\tfrac{12}{25}$ is only the first term, which is what you get by stopping too early.

The first term is $8$, so the sum formula gives an equation in $r$.

$\frac{8}{1-r} = 6 \;\Longrightarrow\; 1 - r = \frac86 = \frac43 \;\Longrightarrow\; r = 1 - \frac43 = -\frac13$

The sign is no surprise: the total $6$ is smaller than the first term $8$, so the later terms must be subtracting on balance, which requires a negative ratio. Its size is $\tfrac13 < 1$, so the series does converge, and the answer is legitimate.

Exceeding $4.99$ means the error is under $0.01$, and the error after $n$ terms is exactly $S\,r^{\,n} = 5(0.6)^n$.

$5(0.6)^n < 0.01 \;\Longleftrightarrow\; (0.6)^n < 0.002$

Take logarithms and divide by $\log(0.6) \approx -0.2218$, reversing the inequality because that number is negative.

$n > \frac{\log(0.002)}{\log(0.6)} = \frac{-2.699}{-0.2218} \approx 12.17$

So $n = 13$ terms are needed. Check the boundary: $5(0.6)^{12} \approx 0.0109$, which leaves the partial sum at about $4.989$, a whisker short, while $5(0.6)^{13} \approx 0.0065$ clears the bar.

Two powers with the same exponent combine into one, since $x^{\,k}y^{\,k} = (xy)^{\,k}$.

$\left(\tfrac12\right)^{\,n-1}\left(\tfrac13\right)^{\,n-1} = \left(\tfrac16\right)^{\,n-1}$

So this is an ordinary geometric series with $a = 1$ and $r = \tfrac16$, which converges.

$S = \frac{1}{1 - \tfrac16} = \frac{1}{\tfrac56} = \frac65$

The two series on their own sum to $2$ and to $\tfrac32$, and the answer is neither their sum $\tfrac72$ nor their product $3$. The terms were multiplied one at a time, which is a different thing from multiplying the two totals, so combine the ratios first and sum once.