Question 1

The table shows the numbers $a _ { n }$ (in thousands)of residents from 2002 through 2007. $$\begin{array} { | c | l | } 
\hline  & \text { Number of residents, } \
&.\
{ \text { Year } }& a _ { n } \
&.\
\hline 2002 & 646 \
\hline 2003 & 657 \
\hline 2004 & 668 \
\hline 2005 & 679 \
\hline
\end{array}$$ $$\begin{array} { | l | l | } 
\hline 2006 & 690 \
\hline 2007 & 701 \
\hline
\end{array}$$
Determine whether a linear model can be used to approximate the data.
If so,find a model algebraically.Let n represent the year,with $n = 2$ corresponding to 2002.
A)A linear model can be used. $a _ { n } = 11 n + 690$
B)A linear model can be used. $a _ { n } = 11 n + 657$
C)A linear model can be used. $a _ { n } = 11 n + 701$
D)A linear model can be used. $a _ { n } = 11 n + 624$
E)A linear model can be used. $a _ { n } = 11 n + 646$

Accepted Answer

The data shows a consistent increase of 11,000 residents each year. To find the model, we can use the year 2002 as a reference point where $n = 2$. For 2002, $a_n = 646$, so substituting $n = 2$ into the equation $a_n = 11n + b$ gives us $646 = 22 + b$, solving for $b$ gives $b = 624$. Therefore, the model is $a_n = 11n + 624$.

Question 2

Find the sum using the formulas for the sums of powers of integers. $\sum _ { n = 1 } ^ { 9 } n ^ { 3 }$
A)729
B)285
C)4050
D)1296
E)2025

Accepted Answer

The formula for the sum of the cubes of the first n natural numbers is $\frac{n^2(n+1)^2}{4}$. Plugging in $n = 9$, we get $\frac{9^2(10)^2}{4} = \frac{81 \cdot 100}{4} = 2025$.

Question 3

Find Pk₊₁ for the given P_k. $$P _ { k } = \frac { 2 } { k ( k + 1 ) }$$ A) $$P _ { k + 1 } = \frac { 2 } { k ( k + 2 ) }$$ B) $$P _ { k + 1 } = \frac { 2 } { k ( k + 1 ) } + \frac { 2 } { ( k + 1 ) ( k + 2 ) }$$ C) $P _ { k + 1 } = \frac { 4 } { ( k + 1 ) ( k + 2 ) }$ D) $P _ { k + 1 } = \frac { 2 } { ( k + 1 ) ( k + 2 ) }$ E) $P _ { k + 1 } = \frac { 2 } { k ( k + 1 ) } + 1$

Accepted Answer

The expression for $P_{k+1}$ is derived by replacing $k$ with $k+1$ in the original formula, resulting in $\frac{2}{(k+1)((k+1)+1)} = \frac{2}{(k+1)(k+2)}$.

Question 4

Use mathematical induction to solve for all positive integers n.​ $3 + 8 + 13 + 18 + \ldots + ( 5 n - 2 ) = ?$ ​
A) $\frac { n } { 6 } ( 5 n + 1 )$
B) $n ( 5 n + 1 )$
C) $\frac { n } { 2 } ( 5 n + 1 )$
D) $\frac { n } { 4 } ( 5 n + 1 )$
E) $5 n + 1$

Accepted Answer

First we must prove the base case:
If n = 1, then 3 = $\frac{1}{2}(5 * 1 + 1)$
This is true, so we move on to the inductive hypothesis:
Assume that for some positive integer k,
$3 + 8 + 13 + 18 + \ldots + (5k - 2) = \frac{k}{2}(5k+1)$
Now we must prove the inductive step, which is to show that this formula holds true for k+1:
$3 + 8 + 13 + 18 + \ldots + (5k - 2) + (5(k+1) - 2) = \frac{(k+1)}{2}(5(k+1) + 1)$
Using the formula for k, we can make the left side look like:
$\frac{k}{2}(5k+1) + (5(k+1) - 2)$
This simplifies to:
$\frac{5}{2}k^2 + \frac{15}{2}k + 3$
We want to rewrite the right side in order to compare it to this expression:
$\frac{(k+1)}{2}(5(k+1) + 1) = \frac{5}{2}k^2 + \frac{15}{2}k + 3$
Therefore, the formula is true for all positive integers n.

Question 5

Find a formula for the sum of the n terms of the sequence. $$\frac { 1 } { 2 } , \frac { 5 } { 4 } , \frac { 25 } { 8 } , \frac { 125 } { 16 } , \ldots$$
A) $\frac { 5 \left( 5 ^ { n } - 2 ^ { n } \right) } { 3 \left( 2 ^ { n } \right) }$
B)​ $\frac { 5 ^ { n - 1 } } { 2 }$
C) $\frac { 5 ^ { n } + 2 ^ { n } } { 7 \left( 2 ^ { n } \right) }$
D) $\frac { 1 } { 2 ^ { n } }$
E) $\frac { 5 ^ { n } - 2 ^ { n } } { 3 \left( 2 ^ { n } \right) }$

Accepted Answer

The answer of Find a formula for the sum of...

Question 6

Find a quadratic model for the sequence with the indicated terms. $a _ { 0 } = 5 , a _ { 2 } = 5 , a _ { 5 } = 65$
A) $a _ { n } = 4 n ^ { 2 } + 5$
B) $a _ { n } = 8 n + 5$
C) $a _ { n } = 4 n ^ { 2 } - 8 n - 5$
D) $a _ { n } = 4 n ^ { 2 } - 8 n + 5$

Accepted Answer

The answer of Find a quadratic model for the sequence...

Question 7

Use mathematical induction to solve for all positive integers n.​ $\sum _ { i = 1 } ^ { n } i ( i + 4 ) = ?$ ​
A) $\frac { n ( 2 n + 13 ) } { 6 }$
B) $\frac { n ( n + 1 ) ( 2 n + 13 ) } { 4 }$
C) $\frac { n ( n + 1 ) ( n + 13 ) } { 6 }$
D) $\frac { n ( n + 1 ) ( 2 n + 13 ) } { 6 }$
E) $\frac { n ( n + 1 ) ( 2 n + 13 ) } { 2 }$

Accepted Answer

We will use mathematical induction to prove that $\sum_{i=1}^ni(i+4)=\frac{n(n+1)(2n+13)}{6}$ holds for all positive integers n.
Base Case: When n = 1, the left hand side is 1(1+4) = 5, and the right hand side is $\frac{1(1+1)(2\cdot 1+13)}{6}$ = 5. Thus, the formula holds for n = 1.
Inductive Step: Assume that the formula holds for some positive integer k, which means that $\sum_{i=1}^k i(i+4) = \frac{k(k+1)(2k+13)}{6}$. We must show that it also holds for k+1, which means that $\sum_{i=1}^{k+1} i(i+4) = \frac{(k+1)(k+2)(2k+15)}{6}$.
We can start by expanding the left hand side:
\begin{align*}
\sum_{i=1}^{k+1} i(i+4) &= (k+1)(k+5) + \sum_{i=1}^{k} i(i+4) \
&= k^2 + 6k + 5 + \frac{k(k+1)(2k+13)}{6} \
&= \frac{6k^2 + 36k + 30 + k^2 + k(2k+13)}{6} \
&= \frac{(k+1)(k+2)(2k+15)}{6}
\end{align*}
where we used the formula for the sum of the first k integers and the induction hypothesis. Therefore, the formula holds for k+1 as well.
By the Principle of Mathematical Induction, the formula holds for all positive integers n. The choice that matches the formula we derived is D.

Question 8

Find the sum using the formulas for the sums of powers of integers. $\sum _ { n = 1 } ^ { 12 } 9 n - 3 n ^ { 2 }$
A)-924
B)-1248
C)468
D)-324
E)-3744

Accepted Answer

The given expression can be split into two separate sums: $9 \sum_{n=1}^{12} n - 3 \sum_{n=1}^{12} n^2$. The sum of the first $n$ integers is $\frac{n(n+1)}{2}$, and the sum of the squares of the first $n$ integers is $\frac{n(n+1)(2n+1)}{6}$. Applying these formulas for $n=12$, we get $9 \times \frac{12 \times 13}{2} - 3 \times \frac{12 \times 13 \times 25}{6} = 9 \times 78 - 3 \times 650 = 702 - 1950 = -1248$.

Quiz 55: Mathematical Induction

Find the sum using the formulas for the sums of powers of integers. $\sum _ { n = 1 } ^ { 9 } n ^ { 3 }$

Find Pk₊₁ for the given P_k. $P _ { k } = \frac { 2 } { k ( k + 1 ) }$

Use mathematical induction to solve for all positive integers n. $3 + 8 + 13 + 18 + \ldots + ( 5 n - 2 ) = ?$

Find a formula for the sum of the n terms of the sequence. $\frac { 1 } { 2 } , \frac { 5 } { 4 } , \frac { 25 } { 8 } , \frac { 125 } { 16 } , \ldots$

Find a quadratic model for the sequence with the indicated terms. $a _ { 0 } = 5 , a _ { 2 } = 5 , a _ { 5 } = 65$

Use mathematical induction to solve for all positive integers n. $\sum _ { i = 1 } ^ { n } i ( i + 4 ) = ?$

Find the sum using the formulas for the sums of powers of integers. $\sum _ { n = 1 } ^ { 12 } 9 n - 3 n ^ { 2 }$

Quiz 55: Mathematical Induction

Find the sum using the formulas for the sums of powers of integers. ∑n=19n3\sum _ { n = 1 } ^ { 9 } n ^ { 3 }∑n=19​n3

Find Pk+1 for the given Pk. Pk=2k(k+1) P _ { k } = \frac { 2 } { k ( k + 1 ) }Pk​=k(k+1) 2​

Use mathematical induction to solve for all positive integers n.​ 3+8+13+18+…+(5n−2) =?3 + 8 + 13 + 18 + \ldots + ( 5 n - 2 ) = ?3+8+13+18+…+(5n−2) =? ​

Find a formula for the sum of the n terms of the sequence. 12,54,258,12516,…\frac { 1 } { 2 } , \frac { 5 } { 4 } , \frac { 25 } { 8 } , \frac { 125 } { 16 } , \ldots21​,45​,825​,16125​,…

Find a quadratic model for the sequence with the indicated terms. a0=5,a2=5,a5=65a _ { 0 } = 5 , a _ { 2 } = 5 , a _ { 5 } = 65a0​=5,a2​=5,a5​=65

Use mathematical induction to solve for all positive integers n.​ ∑i=1ni(i+4) =?\sum _ { i = 1 } ^ { n } i ( i + 4 ) = ?∑i=1n​i(i+4) =? ​

Find the sum using the formulas for the sums of powers of integers. ∑n=1129n−3n2\sum _ { n = 1 } ^ { 12 } 9 n - 3 n ^ { 2 }∑n=112​9n−3n2