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@ -2136,45 +2136,269 @@ Q.E.D.
that $x^{k + 1} = x \cdot x^k$ to prove that for every integer $n \geq 1$,
$\dfrac{d(x^n)}{dx} = nx^{n - 1}$.
**Proof (by mathematical induction):**
Let $P(n)$ be the equation:
$$ \frac{d(x^n)}{dx} = nx^{n - 1} $$
_Basis Step:_
Prove $P(1)$. That is:
$$ \frac{d(x^{(1)})}{dx} = (1)x^{(1) - 1} $$
Alternatively:
$$ \frac{dx}{dx} = 1x^0 $$
Evaluate the left-hand side when $n = 1$:
$$ \frac{dx}{dx} $$
By the given fact that $\dfrac{dx}{dx} = 1$:
$$ = 1 $$
Evaluate the right-hand side when $n = 1$:
$$ = 1x^0 $$
$$ = 1 $$
Both the left and right hand sides of $P(1)$ are equal. Therefore $P(1)$ is
true.
_Inductive Step:_
Let $k$ be any integer where $k \geq 1$.
Suppose $P(k)$. That is:
$$ \frac{d(x^k)}{dx} = kx^{k - 1} $$
This is the inductive hypothesis.
Prove $P(k + 1)$. That is:
$$ \frac{d(x^{(k + 1)})}{dx} = (k + 1)x^{(k + 1) - 1} $$
Alternatively:
$$ \frac{d(x^{(k + 1)})}{dx} = (k + 1)x^k $$
Evaluate left-hand side:
$$ \frac{d(x^{(k + 1)})}{dx} $$
$$ \frac{d(x \cdot x^k)}{dx} $$
By the product rule, we can separate this out into:
$$ \frac{dx}{dx} \cdot x^k + x \cdot \dfrac{d(x^k)}{dx} $$
By the given fact that $\dfrac{dx}{dx} = 1$:
$$ 1 \cdot x^k + x \cdot \dfrac{d(x^k)}{dx} $$
By the inductive hypothesis:
$$ 1 \cdot x^k + x \cdot kx^{k - 1} $$
$$ x^k + x \cdot kx^{k - 1} $$
$$ x^k + kx^{k - 1 + 1} $$
$$ x^k + kx^{k} $$
$$ x^k(1 + k) $$
$$ (k + 1)x^k $$
Evaluate right-hand side:
$$ (k + 1)x^k $$
Both the left and right sides of $P(k + 1)$ are equal. Therefore $P(k + 1)$ is
true.
Q.E.D.
Use the formula for the sum of the first $n$ integers and/or the formula for the
sum of a geometric sequence to evaluate the sums in 20-29 or to write them in
closed form.
20. $4 + 8 + 12 + 16 + \dots + 200$
$$ 4 + 8 + 12 + 16 + \dots + 200 $$
$$ = 4(1 + 2 + 3 + 4 + \dots + 50) $$
$$ = 4\frac{50(51)}{2} $$
$$ = 5100 $$
21. $5 + 10 + 15 + 20 + \dots + 300$
$$ 5 + 10 + 15 + 20 + \dots + 300 $$
$$ = 5(1 + 2 + 3 + 4 + \dots 60) $$
$$ = 5\left(\frac{(60)(61)}{2}\right) $$
$$ = 9150 $$
22.
a. $3 + 4 + 5+ 6 + \dots + 1000$
a. $3 + 4 + 5 + 6 + \dots + 1000$
$$ 3 + 4 + 5 + 6 + \dots + 1000 $$
$$ = (1 + 2 + 3 + 4 + \dots + 1000) - (1 + 2) $$
$$ = \left(\frac{(1000)(1001)}{2}\right) - 3 $$
$$ = 500497 $$
b. $3 + 4 + 5 + 6 + \dots + m$
$$ 3 + 4 + 5 + 6 + \dots + m $$
$$ = (1 + 2 + 3 + 4+ \dots + m) - (1 + 2) $$
$$ = \left(\frac{(m)(m + 1)}{2}\right) - 3 $$
$$ = \frac{m^2 + m}{2} - 3 $$
$$ = \frac{m^2 + m}{2} - \frac{6}{2} $$
$$ = \frac{m^2 + m - 6}{2} $$
23.
a. $7 + 8 + 9 + 10 + \dots + 600$
$$ 7 + 8 + 9 + 10 + \dots + 600 $$
$$ = (1 + 2 + 3 + 4 + \dots + 600) - (1 + 2 + 3 + 4 + 5 + 6) $$
$$ = \left(\frac{(600)(601)}{2}\right) - 21 $$
$$ = 180279 $$
b. $7 + 8 + 9 + 10 + \dots + k$
$$ 7 + 8 + 9 + 10 + \dots + k $$
$$ = (1 + 2 + 3 + 4 + \dots + k) - (1 + 2 + 3 + 4 + 5 + 6) $$
$$ = \left(\frac{(k)(k + 1)}{2}\right) - 21 $$
$$ = \frac{k^2 + k}{2} - 21 $$
$$ = \frac{k^2 + k - 42}{2} $$
24. $1 + 2 + 3 + \dots + (k - 1)$, where $k$ is any integer with $k \geq 2$.
$$ 1 + 2 + 3 + \dots + (k - 1) $$
$$ = \frac{(k - 1)((k - 1) + 1)}{2} $$
$$ = \frac{(k - 1)(k)}{2} $$
$$ = \frac{k^2 - k}{2} $$
25.
a. $1 + 2 + 2^2 + \dots + 2^{25}$
$$ 1 + 2 + 2^2 + \dots + 2^{25} $$
$$ = \frac{2^{25 + 1} - 1}{2^{25} - 1} $$
$$ = \frac{2^{26} - 1}{2 - 1} $$
$$ = 67108863 $$
b. $2 + 2^2 + 2^3 + \dots + 2^{26}$
$$ 2 + 2^2 + 2^3 + \dots + 2^{26} $$
$$ k 2(1 + 2 + 2^2 + \dots + 2^{25}) $$
By part a:
$$ = 2(67108863) $$
$$ = 134217726 $$
c. $2 + 2^2 + 2^3 + \dots + 2^n$
$$ 2 + 2^2 + 2^3 + \dots + 2^n $$
$$ 2(1 + 2 + 2^2 + \dots + 2^{n - 1}) $$
$$ 2\left(\frac{2^{(n - 1) + 1} - 1}{2 - 1}\right) $$
$$ 2\left(\frac{2^n - 1}{1}\right) $$
$$ 2(2^n - 1) $$
$$ 2^{n + 1} - 2 $$
26. $3 + 3^2 + 3^3 + \dots + 3^n$, where $n$ is any integer with $n \geq 1$.
$$ 3 + 3^2 + 3^3 + \dots + 3^n $$
$$ 3(1 + 3 + 3^2 + \dots + 3^{n - 1}) $$
$$ 3\left(\frac{3^{(n - 1) + 1} - 1}{3 - 1}\right) $$
$$ 3\left(\frac{3^n - 1}{2}\right) $$
$$ \frac{3^{n + 1} - 3}{2} $$
27. $5^3 + 5^4 + 5^5 + \dots + 5^k$, where $k$ is any integer with $k \geq 3$.
$$ 5^3 + 5^4 + 5^5 + \dots + 5^k $$
$$ = 5^3(1 + 5 + 5^2 + \dots + 5^{k - 3}) $$
$$ = 5^3\left(\frac{5^{(k - 3) + 1} - 1}{5 - 1}\right) $$
$$ = 5^3\left(\frac{5^{k - 2} - 1}{4}\right) $$
$$ = \frac{5^{k - 2 + 3} - 5^3}{4} $$
$$ = \frac{5^k - 5^3}{4} $$
28. $1 + \dfrac{1}{2} + \dfrac{1}{2^2} + \dots + \dfrac{1}{2^n}$, where $n$ is
any positive integer.
$$ 1 + \dfrac{1}{2} + \dfrac{1}{2^2} + \dots + \dfrac{1}{2^n} $$
$$ = \frac{\left(\dfrac{1}{2}\right)^{n + 1} - 1}{\dfrac{1}{2} - 1} $$
$$ = \frac{\left(\dfrac{1}{2}\right)^{n + 1} - 1}{-\dfrac{1}{2}} $$
$$ = \left[\left(\dfrac{1}{2}\right)^{n + 1} - 1\right](-2) $$
$$ = \left(\dfrac{-2}{2}\right)^{n + 1} + 2 $$
$$ = 2 - \left(\dfrac{-2}{2}\right)^{n + 1} $$
$$ = 2 + \dfrac{1}{2^n} $$
29. $1 - 2 + 2^2 - 2^3 + \dots + (-1)^n2^n$, where $n$ is any positive integer.
$$ 1 - 2 + 2^2 - 2^3 + \dots + (-1)^n2^n $$
$$ = 1 + (-2) + (-2)^2 + (-2)^3 + \dots + (-2)^n $$
$$ = \frac{(-2)^{n + 1} - 1}{(-2) - 1} $$
$$ = \frac{(-2)^{n + 1} - 1}{-3} $$
30. Observe that
$$ \frac{1}{1 \cdot 3} = \frac{1}{3} $$
@ -2187,6 +2411,92 @@ $$ \frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \frac{1}{7
Guess a general formula and prove it by mathematical induction.
General formula:
$$ \frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \dots + \frac{1}{(2n - 1)(2n + 1)} = \frac{n}{2n + 1} $$
for all integers $n \geq 1$.
**Proof (by mathematical induction):**
Let $P(n)$ be the equation:
$$ \frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \dots + \frac{1}{(2n - 1)(2n + 1)} = \frac{n}{2n + 1} $$
_Basis Step:_
Prove $P(1)$:
$$ \frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \dots + \frac{1}{(2(1) - 1)(2(1) + 1)} = \frac{(1)}{2(1) + 1} $$
Evaluate left-hand side when $n = 1$:
$$ \frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \dots + \frac{1}{(2(1) - 1)(2(1) + 1)} $$
$$ = \frac{1}{(2 - 1)(2 + 1)}$$
$$ = \frac{1}{(1)(3)}$$
$$ = \frac{1}{3} $$
Evaluate right-hand side when $n = 1$:
$$ \frac{(1)}{2(1) + 1} $$
$$ \frac{1}{2 + 1} $$
$$ \frac{1}{3} $$
The left and right hand sides of $P(1)$ are equal. Therefore $P(1)$ is true.
_Inductive Step:_
Let $k$ be any integer where $k \geq 1$.
Suppose $P(k)$. That is:
$$ \frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \dots + \frac{1}{(2k - 1)(2k + 1)} = \frac{k}{2k + 1} $$
This is the inductive hypothesis.
Prove $P(k + 1)$. That is:
$$ \frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \dots + \frac{1}{(2(k + 1) - 1)(2(k + 1) + 1)} = \frac{(k + 1)}{2(k + 1) + 1} $$
Alternatively:
$$ \frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \dots + \frac{1}{(2k + 2 - 1)(2k + 2 + 1)} = \frac{k + 1}{2k + 2 + 1} $$
$$ \frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \dots + \frac{1}{(2k + 1)(2k + 3)} = \frac{k + 1}{2k + 3} $$
Evaluate the left-hand side:
$$ \frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \dots + \frac{1}{(2k + 1)(2k + 3)} $$
$$ = \left[\frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \dots + \frac{1}{(2k - 1)(2k + 1)}\right] + \frac{1}{(2k + 1)(2k + 3)} $$
By the inductive hypothesis:
$$ = \frac{k}{2k + 1} + \frac{1}{(2k + 1)(2k + 3)} $$
$$ = \frac{k(2k + 3)}{(2k + 1)(2k + 3)} + \frac{1}{(2k + 1)(2k + 3)} $$
$$ = \frac{k(2k + 3) + 1}{(2k + 1)(2k + 3)} $$
$$ = \frac{2k^2 + 3k + 1}{(2k + 1)(2k + 3)} $$
$$ = \frac{(2k + 1)(k + 1)}{(2k + 1)(2k + 3)} $$
$$ = \frac{k + 1}{2k + 3} $$
Evaluate the right-hand side:
$$ \frac{k + 1}{2k + 3} $$
Both sides of $P(k + 1)$ are equal. Therefore $P(k + 1)$ is true.
Q.E.D.
31. Compute values of the product
$$ \left(1 + \frac{1}{1}\right)\left(1 + \frac{1}{2}\right)\left(1 + \frac{1}{3}\right) \dots \left(1 + \frac{1}{n}\right) $$
@ -2208,11 +2518,109 @@ $$ 1 - 4 + 9 - 16 + 25 = 1 + 2 + 3 + 4 + 5 $$
Guess a general formula and prove it by mathematical induction.
$$ 1 - 4 + 9 - 16 + \dots + (-1)^{n - 1}(n^2) = (-1)^{n - 1}(1 + 2 + 3 + 4 + \dots + n) $$
**Proof (by mathematical induction):**
Let $P(n)$ be the equation:
$$ 1 - 4 + 9 - 16 + \dots + (-1)^{n - 1}(n^2) = (-1)^{n - 1}(1 + 2 + 3 + 4 + \dots + n) $$
for all integers $n \geq 1$.
_Basis Step_:
Prove $P(1)$. That is:
$$ 1 - 4 + 9 - 16 + \dots + (-1)^{(1) - 1}((1)^2) = (-1)^{(1) - 1}(1 + 2 + 3 + 4 + \dots + (1)) $$
Evaluate left-hand side when $n = 1$:
$$ 1 - 4 + 9 - 16 + \dots + (-1)^{(1) - 1}((1)^2) $$
$$ = (-1)^{0}(1^2) $$
$$ = 1(1) $$
$$ = 1 $$
Evaluate right-hand side when $n = 1$:
$$ (-1)^{(1) - 1}(1 + 2 + 3 + 4 + \dots + (1)) $$
$$ = 1 $$
Both sides of $P(1)$ are equal. Therefore $P(1)$ is true.
_Inductive Step_:
Let $k$ be any integer where $k \geq 1$.
Suppose $P(k)$. That is:
$$ 1 - 4 + 9 - 16 + \dots + (-1)^{k - 1}(k^2) = (-1)^{k - 1}(1 + 2 + 3 + 4 + \dots + k) $$
This is the inductive hypothesis.
Prove $P(k + 1)$. That is:
$$ 1 - 4 + 9 - 16 + \dots + (-1)^{(k + 1) - 1}((k + 1)^2) = (-1)^{(k + 1) - 1}(1 + 2 + 3 + 4 + \dots + (k + 1)) $$
Alternatively:
$$ 1 - 4 + 9 - 16 + \dots + (-1)^{k}((k + 1)^2) = (-1)^{k}(1 + 2 + 3 + 4 + \dots + (k + 1)) $$
Evaluate left-hand:
$$ 1 - 4 + 9 - 16 + \dots + (-1)^{k}((k + 1)^2) $$
$$ = \left[1 - 4 + 9 - 16 + \dots + (-1)^{k - 1}(k^2)\right] + (-1)^{k}((k + 1)^2) $$
By the inductive hypothesis:
$$ = (-1)^{k - 1}(1 + 2 + 3 + 4 + \dots + k) + (-1)^{k}((k + 1)^2) $$
$$ = (-1)^{k - 1}\left[(1 + 2 + 3 + 4 + \dots + k) - (k + 1)^2\right] $$
By 5.2.1:
$$ = (-1)^{k - 1}\left[\frac{k(k + 1)}{2} - (k + 1)^2\right] $$
$$ = (-1)^{k - 1}\left[(k + 1)\left(\frac{k}{2} - (k + 1)\right)\right] $$
$$ = (-1)^{k - 1}\left[(k + 1)\left(\frac{k}{2} - \frac{2(k + 1)}{2}\right)\right] $$
$$ = (-1)^{k - 1}\left[(k + 1)\left(\frac{k - 2(k + 1)}{2}\right)\right] $$
$$ = (-1)^{k - 1}\left[(k + 1)\left(\frac{k - 2k - 2)}{2}\right)\right] $$
$$ = (-1)^{k - 1}\left[(k + 1)\left(\frac{-k - 2)}{2}\right)\right] $$
$$ = (-1)^{k - 1}\left[(-1)(k + 1)\left(\frac{k + 2}{2}\right)\right] $$
$$ = (-1)^{k - 1}\left[(-1)\left(\frac{(k + 1)(k + 2)}{2}\right)\right] $$
$$ = (-1)^{k}\left(\frac{(k + 1)(k + 2)}{2}\right) $$
By 5.2.1:
$$ = (-1)^{k}(1 + 2 + 3 + 4 + \dots + (k + 1)) $$
Evaluate right-hand:
$$ (-1)^{k}(1 + 2 + 3 + 4 + \dots + (k + 1)) $$
Both sides of $P(k + 1)$ are equal. Therefore $P(k + 1)$ is true.
Q.E.D.
33. Find a formula in $n$, $a$, $m$, and $d$ for the sum
$(a + md) + (a + (m + 1)d) + (a + (m + 2)d) + \dots + (a + (m + n)d)$, where
$m$ and $n$ are integers, $n \geq 0$, and $a$ and $d$ are real numbers.
Justify your answer.
$$ a + (a + d) + (a + 2d) + \dots (a + nd) = (n + 1)a + d\left(\frac{n(n + 1)}{2}\right) $$
34. Find a formula in $a$, $r$, $m$, and $n$ for the sum
$$ ar^m + ar^{m + 1} + ar^{m + 2} + \dots + ar^{m + n} $$
@ -2220,6 +2628,16 @@ $$ ar^m + ar^{m + 1} + ar^{m + 2} + \dots + ar^{m + n} $$
where $m$ and $n$ are integers, $n \geq 0$, and $a$ and $r$ are real numbers.
Justify your answer.
$$ ar^m + ar^{m + 1} + ar^{m + 2} + \dots + ar^{m + n} = ar^m\left(\frac{r^{n + 1} - 1}{r - 1}\right) $$
By factoring out the $ar^m$, this just becomes a geometric series:
$$ ar^m(1 + r + r^2 + r^3 + \dots r^n) $$
And by 5.2.2, we can substitute that series out with:
$$ ar^m\left(\frac{r^{n + 1} - 1}{r - 1}\right) $$
35. You have two parents, four grandparents, eight great-grandparents, and so
forth.
@ -2227,13 +2645,45 @@ a. If all your ancestors were distinct, what would be the total number of your
ancestors for the past 40 generations (counting your parents' generation as
number one)? (_Hint:_ Use the formula for the sum of a geometric sequence.)
The geometric sequence for this is:
$$ 1 + 2 + 2^2 + 2^3 + \dots + 2^n $$
So, by 5.2.2, this is:
$$ \frac{2^{n + 1} - 1}{2 - 1} $$
Where $n$ is the number of generations. Plugging in 39 (since we count as the
first generation) returns:
$$ \frac{2^{39 + 1} - 1}{2 - 1} $$
$$ = \frac{2^{40} - 1}{1} $$
$$ = 2^{40} - 1 $$
$$ = 1099511627775 $$
b. Assuming that each generation represents 25 years, how long is 40
generations?
$$ 25 \cdot 1099511627775 $$
$$ \approx 2.748779069 \cdot 10^{13} \text{ years} $$
c. The total number of people who have ever lived is approximately 10 billion,
which equals $10^{10}$ people. Compare this fact with the answer to part (a).
What can you deduce?
When demarcated for easier reading, part a's answer reads as:
$$ = 1,099,511,627,775 $$
Which is 1 trillion, 99 billion, 511 million, 627 thousand, 775 people. Since
this exceeds the approximate total number of people who have ever lived. We can
deduce that some(probably many) of my ancestors must have been related to one
another.
Find the mistakes in the proof fragments in 36-38.
36.
@ -2252,6 +2702,22 @@ inductive step, suppose that $k$ is any integer with $k \geq 1$,
$k^2 = \dfrac{k(k + 1)(2k + 1)}{6}$. We must show that
$(k + 1)^2 = \dfrac{(k + 1)((k + 1) + 1)(2(k + 1) + 1)}{6}$."
In the inductive step, the inductive hypothesis reads:
$$ k^2 = \frac{k(k + 1)(2k + 1)}{6} $$
But it should read:
$$ 1^2 + 2^2 + \dots + k^2 = \frac{k(k + 1)(2k + 1)}{6} $$
This error cascades into their proof, which reads:
$$ (k + 1)^2 = \frac{(k + 1)((k + 1) + 1)(2(k + 1) + 1)}{6} $$
But instead should read:
$$ 1^2 + 2^2 + \dots + (k + 1)^2 = \frac{(k + 1)((k + 1) + 1)(2(k + 1) + 1)}{6} $$
37.
**Theorem:**
@ -2271,6 +2737,11 @@ _Show that $P(0)$ is true:_
The left-hand side of $P(0)$ is $1 + 2 + 2^2 + \dots + 2^0 = 1$ and the
right-hand side is $2^{0 + 1} - 1 = 2 - 1 = 1$ also. So $P(0)$ is true."
The left-hand side evaluation should instead read:
The left-hand side of $P(0)$ is $2^0 = 1$ since when $n = 0$, only the first
term is evaluated..
38.
**Theorem:**
@ -2301,10 +2772,51 @@ $$ 1 = 1 $$
Thus $P(1)$ is true."
The author of this proof fragment incorrectly rewrites the upper limit as $i$
instead of $1$. They write:
$$ \sum_{i = 1}^{i}{i(i!)} = (1 + 1)! - 1 $$
When it should be:
$$ \sum_{i = 1}^{1}{i(i!)} = (1 + 1)! - 1 $$
Then, they should evaluate each side independently, but instead they simply
evaluate each together, which is incorrect. Instead the basis step should be
written as:
Evaluate the left-hand side when $n = 1$:
$$ \sum_{i = 1}^{1}{i(i!)} $$
$$ = 1(1!) $$
$$ = 1(1) $$
$$ = 1 $$
Evaluate the right-hand side when $n = 1$:
$$ (1 + 1)! - 1 $$
$$ = (2)! - 1 $$
$$ = (2 \cdot 1) - 1 $$
$$ = 2 - 1 $$
$$ = 1 $$
Both sides of $P(1)$ are equal. Therefore $P(1)$ is true.
39. Use Theorem 5.2.1 to prove that if $m$ and $n$ are any positive integers and
$m$ is odd, then $\sum_{k = 0}^{m - 1}{(n + k)}$ is divisible by $m$. Does
the conclusion hold if $m$ is even? Justify your answer.
Omitted.
40. Use Theorem 5.2.1 and the result of exercise 10 to prove that if $p$ is any
prime number with $p \geq 5$, then the sum of the squares of any $p$
consecutive integers is divisible by $p$.
Omitted.