# Thread: 2003 AP BC Calculus MC Questions

1. ## 2003 AP BC Calculus MC Questions

My teacher assigned us the multiple choice questions from the 2003 BC Calc Ap test and there are four questions that I don't understand how to do.

10. What is the value of ∑ 2^(n+1) $/ 3^n$ ?
I simplified it to ∑ $2 (2/3)^n$
When I did the sum of the geometric series from there I got 6 but the correct answer is 4.

22. What are all values of p for which the infinite series ∑ $n/(n^p+1)$ converges?
(A) p>0
(B) p>=1
(C) p>1
(D) p>=2
(E) p>2
Correct answer is E but I'm not sure where to even start on this problem.

84. A particle moves in the xy-plane so that its position at any time t is given by $x(t) = t^2$ and $y(t) = sin(4t)$. What is the speed of the particle when t = 3?
I figured that you needed to take the derivative of x(t) and y(t) to get dy/dx and then plug in t = 3 but apparently that isn't right. (The correct answer is 6.884)

Any help is greatly appreciated.

2. Originally Posted by Naples
My teacher assigned us the multiple choice questions from the 2003 BC Calc Ap test and there are four questions that I don't understand how to do.

10. What is the value of ∑ 2^(n+1) $/ 3^n$ ?
I simplified it to ∑ $2 (2/3)^n$
When I did the sum of the geometric series from there I got 6 but the correct answer is 4.

22. What are all values of p for which the infinite series ∑ $n/(n^p+1)$ converges?
(A) p>0
(B) p>=1
(C) p>1
(D) p>=2
(E) p>2
Correct answer is E but I'm not sure where to even start on this problem.

84. A particle moves in the xy-plane so that its position at any time t is given by $x(t) = t^2$ and $y(t) = sin(4t)$. What is the speed of the particle when t = 3?
I figured that you needed to take the derivative of x(t) and y(t) to get dy/dx and then plug in t = 3 but apparently that isn't right. (The correct answer is 6.884)

Any help is greatly appreciated.
10) Where does $n$ start $n = 0$ or $n = 1$ (it makes a difference)

22) Compare with $\sum \frac{n}{n^p}$ or $\sum \frac{1}{n^{p-1}}$

84) You want to use $\sqrt{x'^2 + y'^2}$

3. What is the indexing starting at?

4. Your problem with number 10 is that the indexing started at 1 not 0 so when you change your indexing you will obtain 4 as the answer.

5. On number 10 is starts at n=1, but I see what you mean, so thank you.

6. Originally Posted by Naples
On number 10 is starts at n=1, but I see what you mean, so thank you.
$\sum\limits_{n = 1}^\infty {\frac{{2^{n + 1} }}
{{3^n }}} = \frac{{\frac{4}
{3}}}
{{1 - \frac{2}
{3}}}
$

7. How does a = 4/3? This is what I'm getting:

$\sum\limits_{n = 1}^\infty { \frac{2^{n+1}}{3^n} }$

$= 2 \frac{2^n}{3^n} = 2 (\frac{2}{3})^n$

a = 2 right?

8. Originally Posted by lilaziz1
How does a = 4/3? This is what I'm getting:

$\sum\limits_{n = 1}^\infty { \frac{2^{n+1}}{3^n} }$

$= 2 \frac{2^n}{3^n} = 2 (\frac{2}{3})^n$

a = 2 right?
Write out the first few terms

$
\frac{2^2}{3} + \frac{2^3}{3^2} + \frac{2^4}{3^3} + \cdots,
$
$a$ is the first term.

9. Please allow me to be brutally honest.
In any geometric series this is the fact: $\sum\limits_{k = j}^\infty {ax^k } = \frac{{ax^j }}{{1 - x}}$.
That is in words: the sum equals the first term divided by one minus the common ratio.

Can you apply that very simple rule?

10. ah alright. gotcha danny. thx