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A)

There exist real numbers A and B so that \(\frac{1}{k(k + 3)} = \frac{A}{k} + \frac{B}{k + 3}.\) for all real numbers k other than 0 and -3. Enter the ordered pair (A,B).

B)

What is \(\sum_{k=1}^{\infty} \frac 1{k(k+3)}? \)

 Jul 2, 2019
 #1
avatar+979 
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Ok, rethinking this: for A)

try plugging number to find a pattern...

 

For b) you should be able to solve it with a simple knowledge of Sigma notation, try going to khan academy for review.

(maybe somebody else on the forum will help with a fully detailed answer.)

 

-deep regrets,

\(tommarvoloriddle\)

 Jul 2, 2019
edited by tommarvoloriddle  Jul 2, 2019
edited by tommarvoloriddle  Jul 2, 2019
edited by tommarvoloriddle  Jul 2, 2019
 #2
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How do I pay?  I don’t see any links for making a payment.

Are you sure you know how to do this problem. You don’t seem like you know this kind of math.

Guest Jul 2, 2019
 #3
avatar+979 
0

You are supposed to pay? Where?

tommarvoloriddle  Jul 2, 2019
 #6
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tommarvoloriddle, Before your edit, your post read:

 

Ok, rethinking this: for A)

try plugging number to find a pattern... I am not just going to tell the answer in case of a hw prob, this web is not a hw website where you get your answers free. I am sorry. 

Deep regrets

-tommarvoloriddle

 

Your comment ...

“... I am not just going to tell the answer in case of a hw prob, this web is not a hw website where you get your answers free. ...

 

...implies a payment is necessary.   

 

It also had this comment: 

Putting 2 problems in a post like that suggests to me like you don't have enough energy to make two.

 

You have on more than one occasion presented two questions in the same post.   Did you not have enough energy?

Guest Jul 2, 2019
 #7
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You know, doing math at 3 am kind of tires me out, so yes, i was tired.

tommarvoloriddle  Jul 3, 2019
 #8
avatar+476 
+1

You don't have to pay for khan acdemy, what so ever and actually it's say's at the bottom, that it's very mission is to offer free education to people...

HiylinLink  Jul 5, 2019
 #4
avatar+22855 
+2

A)
There exist real numbers \(A\) and \(B\) so that \(\dfrac{1}{k(k + 3)} = \dfrac{A}{k} + \dfrac{B}{k + 3}\).
for all real numbers \(k\) other than \(0\) and \(-3\). Enter the ordered pair \((A,B)\).

 

\(\begin{array}{|lrcll|} \hline &\mathbf{\dfrac{1}{k(k + 3)}} &=& \mathbf{\dfrac{A}{k} + \dfrac{B}{k + 3}} \quad | \quad \cdot k(k + 3) \\\\ &1 &=& \dfrac{Ak(k + 3)}{k} + \dfrac{Bk(k + 3)}{k + 3} \\\\ &1 &=& A (k + 3) + Bk \\ \hline k=-3 &1 &=& A (-3 + 3) + B(-3) \\ &1 &=& 0 -3B \\ &1 &=& -3B \\ & \mathbf{B} &=& \mathbf{-\dfrac{1}{3}} \\ \hline k=0 &1 &=& A (0 + 3) + B(0) \\ &1 &=& 3A+ 0 \\ &1 &=&3A \\ & \mathbf{A} &=& \mathbf{ \dfrac{1}{3}} \\ \hline \end{array} \)

 

\(\mathbf{(A,B) = \left(\dfrac{1}{3},\ -\dfrac{1}{3} \right) }\)

 

\(\begin{array}{|rcll|} \hline \dfrac{1}{k(k + 3)} &=& \dfrac{\frac{1}{3}}{k} + \dfrac{-\frac{1}{3}}{k + 3} \\ &=& \dfrac{1}{3k} - \dfrac{1}{3(k + 3)} \\ \hline \end{array} \)

 

laugh

 Jul 2, 2019
 #5
avatar+22855 
+3

B)
What is \(\sum \limits_{k=1}^{\infty} \dfrac 1{k(k+3)}\)?

 

\(\begin{array}{|rcll|} \hline && \mathbf{ \sum \limits_{k=1}^{\infty} \dfrac 1{k(k+3)} } \\\\ &=& \sum \limits_{k=1}^{\infty} \left( \dfrac{1}{3k} - \dfrac{1}{3(k + 3)} \right) \\\\ &=& \frac13 \sum \limits_{k=1}^{\infty} \left( \dfrac{1}{ k} - \dfrac{1}{k + 3} \right) \\\\ &=& \frac13\left( \sum \limits_{k=1}^{\infty} \dfrac{1}{ k} -\sum \limits_{k=1}^{\infty} \dfrac{1}{k + 3} \right) \\\\ &=& \frac13\left( \sum \limits_{k=1}^{3} \dfrac{1}{ k} +\sum \limits_{k=4}^{\infty} \dfrac{1}{ k} -\sum \limits_{k=1}^{\infty} \dfrac{1}{k + 3} \right) \\\\ &=& \frac13\left( \sum \limits_{k=1}^{3} \dfrac{1}{ k} +\sum \limits_{k=1}^{\infty} \dfrac{1}{ k+3} -\sum \limits_{k=1}^{\infty} \dfrac{1}{k + 3} \right) \\\\ &=& \frac13 \sum \limits_{k=1}^{3} \dfrac{1}{ k} \\\\ &=& \dfrac13 \left( \dfrac{1}{ 1}+\dfrac{1}{ 2}+\dfrac{1}{ 3}\right) \\\\ &=& \dfrac13 \left( \dfrac{6+3+2}{ 6} \right) \\\\ &=& \dfrac13 \left( \dfrac{11}{ 6} \right) \\\\ &=& \mathbf{ \dfrac{11}{ 18} } \\ \hline \end{array}\)

 

laugh

 Jul 2, 2019

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