summation capabilities

[jkh1919]

Hello, I"m having some trouble with summation capabilities problems. Here are the ones (sorry, E(sigma) is the closest thing I could come up with to the actual sigma notation symbol):

Find the Sum: (for this one I really have no clue how to even start it. The summation formulas I have do not appear to address problems like this, but then again I am the type of person who needs something pointed out to him or I will not notice it. It's not voluntary I promise)

7
E(sigma) 1 / k^2+7
k=3

Find the Sum: (same thing for this one, I know the answer is simple but I still do not get it)

24
E(sigma)4i
i=1

Find the Sum: (I'm guessing the formula for this one would be:

n
E(sigma)i^2 = n(n+1)(2n+1) / 6
i=1

but have tried numerous times and have come up with the wrong answer repeatedly. I'd just really appreciate it if someone could show me the proper way to apply this formula for this problem or tell me if I am even using the right formula for the problem below):

9
E(sigma)(i-1)^2
i=1

I have the summation formulas and have tried applying them the best way I can, but am still coming up with wrong answers. If someone can please show me the steps to working these 3 problems, you'd be helping me out tremendously. I'd greatly appreciate it.

 

[mmm4444bot]

I have … tried …

… coming up with wrong answers …

… please show me the steps to working these 3 problems …

Providing a listing of steps upon request is not how things work, at this help site. We first need to see some of those efforts or an explanation about why/where you are stuck before we are able to determine how to begin guiding you. Please check our FORUM GUIDELINES.

In the meantime, I'll give you a hint.

24
E(sigma)4i
i=1

Σ(4i) is the same as 4Σ(i)

I bet that you can figure out the sum of the first 24 Natural numbers, and then multiply that by 4.

Cheers :cool:

 

[mmm4444bot]

1 / k^2+7

Because of your word-spacing above, I interpret that expression as \(\displaystyle \frac{1}{k^2 + 7}\).

Is that correct? If so, text the denominator with grouping symbols: 1/(k^2 + 7)

Otherwise, some people will interpret your typing as \(\displaystyle \frac{1}{k^2} + 7\)

 

[jkh1919]

Because of your word-spacing above, I interpret that expression as \(\displaystyle \frac{1}{k^2 + 7}\).

Is that correct? If so, text the denominator with grouping symbols: 1/(k^2 + 7)

Otherwise, some people will interpret your typing as \(\displaystyle \frac{1}{k^2} + 7\)

The first one is correct and I'll go and work through the steps for each problem...the problem is when I'm looking at these formulas, none of these problems are set up in a way that I can tell if I am even using the right one. Can someone look at my original post and at least tell me if the last problem I wrote about will need the formula that I referenced?

 

[jkh1919]

I bet that you can figure out the sum of the first 24 Natural numbers, and then multiply that by 4.

Cheers :cool:

Thanks, yes I do believe I can do that. Thanks for the tip that is all I needed is a starting point. I won't go into why I sit through 3 hours of class every day and absorb absolutely zero information, I'll just bite the bullet and accept the fact that I am appearing and coming off as completely lazy and that I am asking others to do my work for me. (which by the way, is not what is actually going on here, I'll just leave it at that).

 

[mmm4444bot]

the problem is when I'm looking at these formulas, none of these problems are set up in a way that I can tell if I am even using the right [formulas].

Sometimes, we need to manipulate the givens into a form where we do recognize a way to proceed.

A famous problem solver (Pólya) once said that problem solving is basically looking at what you know in a way that makes the solution obvious (paraphrased).

In other words, sometimes we need to manipulate or rearrange what we are given. The following is an example of what I'm talking about.

Can someone look at my original post and at least tell me if the last problem I wrote about will need the formula that I referenced?

9
Σ[(i - 1)^2]
1

If you expand the given binomial square, you'll end up with a polynomial; the summation can then be broken-up into summations of each of the three polynomial terms.

Σ[i^2] - Σ[2i] + Σ[1]

Hence, the formula that you edited into your original post is appropriate for the Σ[i^2] piece. :cool:


PS: I did not assume that you are lazy; I was holding off on making a judgment until after I had more information. Cheers

 

[Yogi]

Sometimes, we need to manipulate the givens into a form where we do recognize a way to proceed.
9
Σ[(i - 1)^2]
1

Σ[i^2] - Σ[2i] + Σ[1]

Although this method is useful to know for some problems, it uses unnecessary steps here and takes longer.
The sequence is 0,1,4,...,64 so simply sum the first 8 squares using the formula n(n+1)(2n+1) / 6 and you are done in 1 step instead of 3.

 

[mmm4444bot]

For your third problem, there are TWo ways to solve it. One is the way that mmm already told you. The other is to substitute j = i - 1.

That j-substitution is a formal path to what Yogi posted, yes?

One can certainly evaluate the sequence, recognize the first eight squares, and go from there.

If one completes the steps required to arrive at that recognition, great. If not, they might be able to get away with then simply doing the arithmetic manually and enter the sums into the appropriate field for the entire assignment. (That would constitute a third method, yes?)

At the end of the day, the grade all comes down to how much of a machine their teacher is.

I am concerned that the OP had difficulty with the second exercise. I would like to know more about their situtation, before tutoring further.

 

[mmm4444bot]

simple addition is not suitable for THIS problem because the limit is not specified numerically.

We're talking about the sum where i takes on the first nine Natural numbers, yes?