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 \(\sum_{k=1}^{\infty} \frac 1{k(k+3)}? \)
+1712 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,
\(tommarvoloriddle\)
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.
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?
+1712 You know, doing math at 3 am kind of tires me out, so yes, i was tired.
+434 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...
+26367 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 \((A,B)\).
\(\begin{array}{|lrcll|} \hline &\mathbf{\dfrac{1}{k(k + 3)}} &=& \mathbf{\dfrac{A}{k} + \dfrac{B}{k + 3}} \quad | \quad \cdot k(k + 3) \\\\ &1 &=& \dfrac{Ak(k + 3)}{k} + \dfrac{Bk(k + 3)}{k + 3} \\\\ &1 &=& A (k + 3) + Bk \\ \hline k=-3 &1 &=& A (-3 + 3) + B(-3) \\ &1 &=& 0 -3B \\ &1 &=& -3B \\ & \mathbf{B} &=& \mathbf{-\dfrac{1}{3}} \\ \hline k=0 &1 &=& A (0 + 3) + B(0) \\ &1 &=& 3A+ 0 \\ &1 &=&3A \\ & \mathbf{A} &=& \mathbf{ \dfrac{1}{3}} \\ \hline \end{array} \)
\(\mathbf{(A,B) = \left(\dfrac{1}{3},\ -\dfrac{1}{3} \right) }\)
\(\begin{array}{|rcll|} \hline \dfrac{1}{k(k + 3)} &=& \dfrac{\frac{1}{3}}{k} + \dfrac{-\frac{1}{3}}{k + 3} \\ &=& \dfrac{1}{3k} - \dfrac{1}{3(k + 3)} \\ \hline \end{array} \)
+26367 B)
What is \(\sum \limits_{k=1}^{\infty} \dfrac 1{k(k+3)}\)?
\(\begin{array}{|rcll|} \hline && \mathbf{ \sum \limits_{k=1}^{\infty} \dfrac 1{k(k+3)} } \\\\ &=& \sum \limits_{k=1}^{\infty} \left( \dfrac{1}{3k} - \dfrac{1}{3(k + 3)} \right) \\\\ &=& \frac13 \sum \limits_{k=1}^{\infty} \left( \dfrac{1}{ k} - \dfrac{1}{k + 3} \right) \\\\ &=& \frac13\left( \sum \limits_{k=1}^{\infty} \dfrac{1}{ k} -\sum \limits_{k=1}^{\infty} \dfrac{1}{k + 3} \right) \\\\ &=& \frac13\left( \sum \limits_{k=1}^{3} \dfrac{1}{ k} +\sum \limits_{k=4}^{\infty} \dfrac{1}{ k} -\sum \limits_{k=1}^{\infty} \dfrac{1}{k + 3} \right) \\\\ &=& \frac13\left( \sum \limits_{k=1}^{3} \dfrac{1}{ k} +\sum \limits_{k=1}^{\infty} \dfrac{1}{ k+3} -\sum \limits_{k=1}^{\infty} \dfrac{1}{k + 3} \right) \\\\ &=& \frac13 \sum \limits_{k=1}^{3} \dfrac{1}{ k} \\\\ &=& \dfrac13 \left( \dfrac{1}{ 1}+\dfrac{1}{ 2}+\dfrac{1}{ 3}\right) \\\\ &=& \dfrac13 \left( \dfrac{6+3+2}{ 6} \right) \\\\ &=& \dfrac13 \left( \dfrac{11}{ 6} \right) \\\\ &=& \mathbf{ \dfrac{11}{ 18} } \\ \hline \end{array}\)
