1. ## Evaulating Riemann Summations

$

\sum _{k=7}^{77} (k)sin\frac{(\pi)}{2} + \frac{cos(\pi)}{2}
$

Here is my logic, not sure if its 100% concrete. Let z = $\frac{(k)sin(\pi)}{2} + \frac{cos(\pi)}{2}$

then the first term of k will be 7.. 8 ... 9 .... 77. Obviously summing this up one at a time would be a daunting calculating task which is prone to error. So, im looking for a quicker way to do it. Since we can assume that i is incrementing by one each time and i use the formula n(n+1)/2 and then multiply it by the equation given? If thats the case, i got 3003 when i did that, but this seems like too large of a number to multiply this fraction by. Any suggestions?

2. Originally Posted by tokio
$

\sum _{k=7}^{77} = \frac{(k)sin(\pi)}{2} + \frac{cos(\pi)}{2}
$

Here is my logic, not sure if its 100% concrete. Let z = $\frac{(k)sin(\pi)}{2} + \frac{cos(\pi)}{2}$

then the first term of k will be 7.. 8 ... 9 .... 77. Obviously summing this up one at a time would be a daunting calculating task which is prone to error. So, im looking for a quicker way to do it. Since we can assume that i is incrementing by one each time and i use the formula n(n+1)/2 and then multiply it by the equation given? If thats the case, i got 3003 when i did that, but this seems like too large of a number to multiply this fraction by. Any suggestions?
Check your post. what you wrote makes no sense. (By the way, $\sin \pi = 0 \text{ and } \cos \pi = -1$)

3. Originally Posted by Jhevon
Check your post. what you wrote makes no sense. (By the way, $\sin \pi = 0 \text{ and } \cos \pi = -1$)
Ok, i admit i have no idea what im doing! Im trying to think of a way to figure this out, any suggestions?

4. Originally Posted by tokio
Ok, i admit i have no idea what im doing! Im trying to think of a way to figure this out, any suggestions?
i mean your question, the problem you posted itself does not make sense.

$\sum_{k = 7}^{77}$ makes no sense by itself. You need something after the summation sign.

5. Originally Posted by Jhevon
i mean your question, the problem you posted itself does not make sense.

$\sum_{k = 7}^{77}$ makes no sense by itself. You need something after the summation sign.
Sorry, there is no equal sign.

6. Originally Posted by tokio
$

\sum _{k=7}^{77} \frac{(k)sin(\pi)}{2} + \frac{cos(\pi)}{2}
$

Here is my logic, not sure if its 100% concrete. Let z = $\frac{(k)sin(\pi)}{2} + \frac{cos(\pi)}{2}$

then the first term of k will be 7.. 8 ... 9 .... 77. Obviously summing this up one at a time would be a daunting calculating task which is prone to error. So, im looking for a quicker way to do it. Since we can assume that i is incrementing by one each time and i use the formula n(n+1)/2 and then multiply it by the equation given? If thats the case, i got 3003 when i did that, but this seems like too large of a number to multiply this fraction by. Any suggestions?
As in $\sum_{k = 7}^{77} \frac {k \sin \pi}2 + \frac {\cos \pi}2$ or $\sum_{k = 7}^{77} \left( \frac {k \sin \pi}2 + \frac {\cos \pi}2 \right)$ ?

Since $\sin \pi = 0$ and $\cos \pi = -1$, in the first case, the answer is just -1/2.

in the second case, what you have is $\sum_{k = 7}^{77} (-1/2) = - \frac 12 \sum_{k = 7}^{77}1$, and i suppose you know how to deal with that.

7. Originally Posted by Jhevon
As in $\sum_{k = 7}^{77} \frac {k \sin \pi}2 + \frac {\cos \pi}2$ or $\sum_{k = 7}^{77} \left( \frac {k \sin \pi}2 + \frac {\cos \pi}2 \right)$ ?

Since $\sin \pi = 0$ and $\cos \pi = -1$, in the first case, the answer is just -1/2.

in the second case, what you have is $\sum_{k = 7}^{77} (-1/2) = - \frac 12 \sum_{k = 7}^{77}1$, and i suppose you know how to deal with that.

Tried to edit the original promblem once again with latex, but cant figure it out . The original problem is ksin (pi/2) + cos(pi/2). In other words it looks like its sin(function) as opposed to $\frac{(sin * \pi)}{2}.$. I end up getting 3003 again once i was done.

8. Originally Posted by tokio
Tried to edit the original promblem once again with latex, but cant figure it out . The original problem is ksin (pi/2) + cos(pi/2). In other words it looks like its sin(function) as opposed to $\frac{(sin * \pi)}{2}.$. I end up getting 3003 again once i was done.
*sigh* ok, one more time. (by the way, $\sin * \pi$ makes no sense. you mean $\sin \pi$ or $\sin (\pi)$--not multiplication).

Anyway, $\sin \frac {\pi}2 = 1,~\cos \frac {\pi}2 = 0$, so your series is:

$\sum_{k = 7}^{77} k = \sum_{k = 1}^{77}k - \sum_{k = 1}^6k$

and you know the formula to calculate each of those