1. ## summation

Evaluate exactly:

a. $
\sum_{k=1}^{100} (1 + (-1)^k)
$

b. $
\sum_{k=3}^{6} \frac{k^2+2}{k-2}
$

For these above do I just plug in the values within the interval into k and then add the total?

so for b would it be:

$
\frac{3^2 + 2}{3 - 2} + \frac{4^2 + 2}{4 - 2} + \frac{5^2 + 2}{5 - 2} + \frac{6^2 + 2}{6 - 2} = 11 + 9 + 9 + 9.5 = 38.5$

Is this correct or am I way off? haha

Thank you to anyone who attempts this for me : )

2. $
\sum_{k=1}^{100} (1 + (-1)^k)=0+2+0+2 ... = 100$

As for exercice 2 ... exactly.

3. Originally Posted by qzno
Evaluate exactly:

a. $
\sum_{k=1}^{100} (1 + (-1)^k)
$

b. $
\sum_{k=3}^{6} \frac{k^2+2}{k-2}
$

For these above do I just plug in the values within the interval into k and then add the total?

so for b would it be:

$
\frac{3^2 + 2}{3 - 2} + \frac{4^2 + 2}{4 - 2} + \frac{5^2 + 2}{5 - 2} + \frac{6^2 + 2}{6 - 2} = 11 + 9 + 9 + 9.5 = 38.5$

Is this correct or am I way off? haha

Thank you to anyone who attempts this for me : )
For the first one, write out some of the terms and group in pairs (some will cancel).

The second you have is correct.

4. Thank you both very much : )

5. Hello, qzno!

Evaluate exactly:

$(a)\;\;\sum_{k=1}^{100} (1 + (-1)^k)$

$(b)\;\;\sum_{k=3}^{6} \frac{k^2+2}{k-2}$

Do I just plug in the values into $k$ and then add the total? . . . . Yes!

So for (b) would it be:

$\frac{3^2 + 2}{3 - 2} + \frac{4^2 + 2}{4 - 2} + \frac{5^2 + 2}{5 - 2} + \frac{6^2 + 2}{6 - 2} \:= \:11 + 9 + 9 + 9.5 \:= \:38.5$ . . . . Right!

$\text{For (a), we have: }\; \underbrace{0 + 2 + 0 + 2 + \hdots + 0 + 2}_{\text{100 terms}}$

6. Originally Posted by qzno
Evaluate Exactly

$

\sum_{k=1}^{100} (1 + (-1)^k)

$

I think this can be separated to form

$

\sum_{k=1}^{100} 1 + \sum_{k=1}^{100} (-1)^k

$

But I do not know where to go from here

Are there any rules or formulas to complete this?

And I know that

$

\sum_{k=1}^{100} 1 = 100

$
What is the value of (-1)^k for even values of k? For odd values of k?

What then is the value of 1 + (-1)^k for even values of k? For odd values of k?

How many even-value pairs do you have? How many odd?

What then is the sum?

7. Originally Posted by qzno
Evaluate Exactly

$

\sum_{k=1}^{100} (1 + (-1)^k)

$

I think this can be separated to form

$

\sum_{k=1}^{100} 1 + \sum_{k=1}^{100} (-1)^k

$

But I do not know where to go from here

Are there any rules or formulas to complete this?

And I know that

$

\sum_{k=1}^{100} 1 = 100

$
You're getting there. Try expanding the sums so you get an idea of what's going on:

${\color{red}\sum_{k=1}^{100} 1} + {\color{blue}\sum_{k=1}^{100} (-1)^k} = \underbrace{{\color{red}1 + 1 + \cdots + 1}}_{100 \ \text{times}} \ + \ \left[{\color{blue}(-1) + (-1)^2 + \cdots + (-1)^{99} + (-1)^{100}}\right]$

Recall that: $(-1)^n = \begin{cases} 1 & \text{ if } n \text{ is even} \\ -1 & \text{ if } n \text{ is odd} \end{cases}$

So simplifying the blue:

\begin{aligned} {\color{red}\sum_{k=1}^{100} 1} + {\color{blue}\sum_{k=1}^{100} (-1)^k} & = {\color{red}100} \ + \ \left[{\color{blue}\underbrace{(-1) + 1}_{= 0} + \underbrace{(-1) + 1}_{= 0} + \cdots + \underbrace{(-1) + 1}_{= 0} }\right]\end{aligned}

I'm sure you can take it from here.

8. Originally Posted by qzno
Evaluate Exactly

$\sum_{k=1}^{100} (1 + (-1)^k)$

[snip]
i know that the final answer is 100 overall
because when u put -1 to an even power, it becomes a +1
and when u put -1 to an odd power, it becomes a -1

and you have 50 even and 50 odd numbers between 1 and 100
but i was wondering if there were any formulas or rules for this : )

9. Excellent o_O
Thanks For Responding Both Of You : )