Arithmetic and geometric sequences and sums

Worked solutions

Problem 1

1. $7d = 51 - 23 = 28$, so $d = 4$From term $5$ to term $12$ is $12 - 5 = 7$ steps.
2. $a_1 = 23 - 4\cdot 4 = 7$Walk back four steps from $a_5$.
3. $a_n = 7 + 4(n - 1) = 4n + 3$The $n$-th term formula, simplified.
4. $a_{40} = 4\cdot 40 + 3 = 163$Check the formula: $a_{12} = 48 + 3 = 51$.
5. $d = 4$, $a_1 = 7$, $a_n = 4n + 3$, $a_{40} = 163$An arithmetic sequence is a straight line sampled at whole numbers: $a_n = 4n + 3$ has slope $d = 4$, and any two terms fix the line.

Problem 2

1. First: $7\cdot 15 = 105$; last: $7\cdot 71 = 497$$100/7 \approx 14.3$ and $500/7 \approx 71.4$, so the multiples run from $7\cdot 15$ to $7\cdot 71$.
2. $n = 71 - 15 + 1 = 57$ termsOne term for each of $15, 16, \dots, 71$.
3. $S_{57} = \dfrac{57(105 + 497)}{2}$Arithmetic with $d = 7$: number of terms times the average of first and last.
4. $= 57\cdot 301$$(105 + 497)/2 = 602/2 = 301$.
5. $17\,157$This is Gauss's pairing: $105 + 497$, $112 + 490$, and so on all equal $602$. With an odd number of terms the middle one, $301 = 7\cdot 43$, pairs with itself, and the formula still holds.

Problem 3

1. $a_1 = 3$, $d = 4$, $a_n = 3 + 4(n - 1) = 4n - 1$The common difference is $7 - 3 = 4$.
2. $S_n = \dfrac{n(3 + 4n - 1)}{2} = n(2n + 1)$First plus last is $4n + 2$; halve it.
3. $2n^2 + n - 1000 = 0$ at $n = \dfrac{-1 + \sqrt{8001}}{4}$The quadratic formula, positive root.
4. $n \approx 22.11$$\sqrt{8001} \approx 89.45$.
5. $S_{22} = 22\cdot 45 = 990$; $S_{23} = 23\cdot 47 = 1081$$S_n$ increases with $n$, so check the whole numbers on each side of the root.
6. $S_n = n(2n + 1) = 2n^2 + n$; the smallest $n$ is $23$The sum of an arithmetic sequence is quadratic in $n$, as the area under a line is quadratic in its length. The root $22.11$ must be rounded up: $22$ terms are not enough (Mistake 5).

Problem 4

1. $r^3 = \dfrac{a_6}{a_3} = -8$, so $r = -2$Three steps of the ratio separate term $3$ from term $6$; a cube root keeps the sign.
2. $a_1 = \dfrac{12}{(-2)^2} = 3$$a_3 = a_1r^2$.
3. $S_8 = \dfrac{3(1 - (-2)^8)}{1 - (-2)}$The sum formula with $n = 8$.
4. $= \dfrac{3(1 - 256)}{3} = -255$$(-2)^8 = 256$, and $1 - (-2) = 3$.
5. $r = -2$, $a_1 = 3$, $S_8 = -255$Check by adding: $3 - 6 + 12 - 24 + 48 - 96 + 192 - 384 = -255$. With an even number of terms and $r < -1$, the last, negative term is the largest in size, so the sum is negative.

Problem 5

1. $S_n = a_1 + a_1r + \dots + a_1r^{n-1}$The $n$ terms, from $r^0$ to $r^{n-1}$.
2. $rS_n = a_1r + a_1r^2 + \dots + a_1r^n$Multiply every term by $r$: the list shifts one place.
3. $S_n - rS_n = a_1 - a_1r^n$Every term but the first of $S_n$ and the last of $rS_n$ cancels.
4. $S_n = \dfrac{a_1(1 - r^n)}{1 - r}$Factor out $S_n$ and divide by $1 - r$, which is allowed because $r \ne 1$.
5. Chessboard: $a_1 = 1$, $r = 2$, $n = 64$, so $S_{64} = \dfrac{1 - 2^{64}}{1 - 2} = 2^{64} - 1$$64$ squares hold $2^0$ to $2^{63}$.
6. $S_n = \dfrac{a_1(1 - r^n)}{1 - r}$; the board holds $2^{64} - 1 \approx 1.845\times 10^{19}$ grainsThe same sum is the largest unsigned $64$-bit integer, $18\,446\,744\,073\,709\,551\,615$: in binary it is sixty-four $1$s. For $r > 1$ the form $a_1(r^n - 1)/(r - 1)$ keeps both parts positive.

Problem 6

1. $2.1\overline{45} = 2.1 + 0.045 + 0.00045 + \dots$Split off the part that does not repeat; each repeat of $45$ is two places further right.
2. The tail is geometric with $a_1 = 0.045$ and $r = 0.01$Each term is $\tfrac{1}{100}$ of the one before.
3. Tail $= \dfrac{0.045}{1 - 0.01} = \dfrac{0.045}{0.99}$$\lvert r\rvert < 1$, so the infinite sum is $a_1/(1 - r)$.
4. $= \dfrac{45}{990} = \dfrac{1}{22}$Multiply top and bottom by $1000$, then cancel $45$.
5. $\dfrac{21}{10} + \dfrac{1}{22} = \dfrac{231 + 5}{110} = \dfrac{236}{110}$Common denominator $110$.
6. $\dfrac{118}{55}$Check: $118/55 = 2.14545\ldots$. Every repeating decimal is a geometric series with ratio $10^{-p}$ for a block of $p$ digits, which is why every repeating decimal is a fraction.

Problem 7

1. First sum: $15 - 4 + 1 = 12$ terms, from $2\cdot 4 - 5 = 3$ to $2\cdot 15 - 5 = 25$The terms go up by $2$ each time: arithmetic with $d = 2$.
2. $\sum_{k=4}^{15}(2k - 5) = \dfrac{12(3 + 25)}{2} = 168$Number of terms times the average of first and last.
3. Second sum: $j = k - 2$ runs from $0$ to $5$, and $3\cdot 2^k = 3\cdot 2^{j+2} = 12\cdot 2^j$Substituting $k = j + 2$ moves the shift into the coefficient.
4. $\sum_{j=0}^{5} 12\cdot 2^j = \dfrac{12(2^6 - 1)}{2 - 1}$Geometric with $a_1 = 12$, $r = 2$, and $6$ terms.
5. $= 12\cdot 63 = 756$Check: $12 + 24 + 48 + 96 + 192 + 384 = 756$.
6. $168$, and $\sum_{j=0}^{5} 12\cdot 2^j = 756$Before applying a formula, read off three things from the sigma: the first term, the ratio or difference, and the number of terms, upper minus lower plus one.

Problem 8

1. $v_1 = (1 - \beta)g_1$, $v_2 = (1 - \beta)(\beta g_1 + g_2)$, and so onUnroll the recursion from $v_0 = 0$: each step multiplies the old terms by $\beta$ and adds a new one.
2. $v_t = (1 - \beta)\sum_{k=1}^{t}\beta^{t-k}g_k$By induction: $\beta v_{t-1}$ raises every power by one, and $(1 - \beta)g_t$ is the $k = t$ term with $\beta^0$.
3. Weights: $(1 - \beta)(1 + \beta + \dots + \beta^{t-1})$The powers $\beta^{t-k}$ for $k = t, t-1, \dots, 1$, a geometric sum of $t$ terms.
4. $= (1 - \beta)\cdot\dfrac{1 - \beta^t}{1 - \beta} = 1 - \beta^t$Problem 5's formula with $a_1 = 1$ and $r = \beta$.
5. $v_{10} = (1 - 0.9^{10})g \approx 0.6513\,g$With a constant input, $v_t$ is the total weight times $g$; $0.9^{10} \approx 0.3487$.
6. $\hat v_{10} = \dfrac{0.6513\,g}{0.6513} = g$Dividing by the total weight makes the weights add up to $1$.
7. $0.9^t < 0.01$ when $t > \dfrac{\ln 0.01}{\ln 0.9} \approx 43.7$Take logs; $\ln 0.9 < 0$ flips the inequality.
8. $v_{10} \approx 0.651\,g$, $\hat v_{10} = g$; the weights pass $0.99$ at $t = 44$Starting from $0$ biases the average toward $0$ by the factor $1 - \beta^t$, and Adam's bias correction divides it out. By $t = 44$ the factor is $1 - 0.9^{44} \approx 0.9903$ and the correction barely matters; with $\beta = 0.999$, as for Adam's second moment, it takes about $4600$ steps.

Problem 9

1. $T = x + 2x^2 + 3x^3 + \dots$ and $xT = x^2 + 2x^3 + \dots$Multiply by $x$: each coefficient moves one power up.
2. $T - xT = x + x^2 + x^3 + \dots$Coefficients $k$ and $k - 1$ meet at each power $x^k$, leaving $1$.
3. $= \dfrac{x}{1 - x}$A geometric series with $a_1 = x$, $r = x$ and $\lvert x\rvert < 1$.
4. $T = \dfrac{x}{(1 - x)^2}$Divide by $1 - x$. The series converges for $\lvert x\rvert < 1$, since the ratio of consecutive terms, $\tfrac{k+1}{k}\lvert x\rvert$, tends to $\lvert x\rvert < 1$; so $T$ is a finite number and the subtraction is allowed.
5. $x = \tfrac12$: $T = \dfrac{1/2}{1/4} = 2$Partial sums: $0.5$, $1$, $1.375$, $1.625, \dots$
6. Mean: $\dfrac{p}{1 - p}\sum_{k} k(1 - p)^k = \dfrac{p}{1 - p}\cdot\dfrac{1 - p}{p^2} = \dfrac1p$Take one factor $1 - p$ outside, then step 4 with $x = 1 - p$, $1 - x = p$.
7. $\sum kx^k = \dfrac{x}{(1 - x)^2}$; $\sum k/2^k = 2$; the mean is $1/p = 5$ trialsA sum whose terms are an arithmetic sequence times a geometric one is called arithmetico-geometric, and the subtraction that proves the geometric formula reduces it to a geometric sum. The same answer comes from differentiating $\sum x^k = 1/(1 - x)$ and multiplying by $x$.

Problem 10

1. After $n$ months the debt has grown to $L(1 + i)^n$ with $L = 20\,000$, $i = 0.005$Each month multiplies the balance by $1 + i$.
2. Payment $k$ has grown to $P(1 + i)^{60 - k}$ by month $60$It is made at the end of month $k$ and earns interest for the remaining months.
3. $P\sum_{j=0}^{59}(1 + i)^j = P\cdot\dfrac{(1 + i)^{60} - 1}{i}$A geometric sum with $a_1 = 1$, $r = 1 + i$ and $60$ terms.
4. $L(1 + i)^{60} = P\cdot\dfrac{(1 + i)^{60} - 1}{i}$The loan is repaid when the payments, grown to month $60$, equal the grown debt.
5. $P = \dfrac{Li}{1 - (1 + i)^{-60}} = \dfrac{100}{1 - 1.005^{-60}}$Divide both sides by $(1 + i)^{60}$; $Li = 100$.
6. $1.005^{60} \approx 1.34885$, so $1 - 1.005^{-60} \approx 0.25863$One power, then its reciprocal.
7. Payment $\approx 386.66$ dollars; total paid $\approx 23\,199.36$, so interest $\approx 3\,199.36$ dollars$60\times 386.656 \approx 23\,199.36$. Simple interest on the full $20\,000$ for five years would be $6\,000$; it is about half that because the balance falls as it is repaid.