Question

These are 2 out of 6 questions from my assignment. I actually need to know how to do them rather than just being given the answers as I .. which is rare today.. want to learn.

Write the sum using summation notation, assuming the suggested pattern continues.1 - 3 + 9 - 27 + ...

Answer 1

Write the sum using summation notation, assuming the suggested pattern continues.

25 + 36 + 49 + 64 + ... + n2 + ...

Answer 2

Well, I don't know any straightforward of clear-cut way to do this, but do you see a pattern with
1, -3, 9, -27...etc?

Answer 3

\[\sum_{n=5}^{∞} n^2\] <-- The equation is suppose to come out with an answer like that.. But I don't know the exact answer.

Answer 4

This is for the second, isn't it? Well, just notice the pattern...
25 = 5^2
35 = 6^2
49 = 7^2

So it's just a series of squares, starting at 5 :)

Answer 5

Soo.. Exactly how do I put this as an answer? Because when I look at the Sigma Sign I just get confused.

Answer 6

Just a quick question... do you code (program)?

Answer 7

Yes. C++, C#, PHP, DM, Java, etc. Why?

Answer 8

Well, you can think of that sigma as a loop command :)
\[\huge \sum_{n=5}^{k}n^2\]
is like

----
int n = 5
int s = 0
while(n <= k)
{
s = s+(n^2)
n = n +1
}

//LOL

Answer 9

Does that help? :D

Answer 10

A little bit. So.. while the number the right of the Sigma is less than or equal to the number up top... where does the integer S come from? lol

Answer 11

s it the sum.
Total sum

Forgot the
return s;
part :)

Answer 12

Because the sigma is like a loop function that returns s.

Answer 13

or rather, returns the sum :)

Answer 14

So, for instance
\[\huge \sum_{n=5}^{9}n^2=5^2+6^2+7^2+8^2+9^2\]and stop at 9 :)

Answer 15

hey, @KrisStudy
It's not the number to the right of sigma, but the number below it, the n.
While it is still less than or equal to, in this case, 9, then you just keep adding...

Answer 16

Okay. I understand. :) Gotcha. Let me try and see if I can get this right.

Answer 17

\[\sum_{n=5}^{infinity}n^2\] & \[\sum_{n=0}^{infinity}1(-3)^n\] Are the answers to those two? Do you know?

Answer 18

Yep. Although, the first one is the answer to your second series, and vice versa.
By the way... infinity symbol = \infty

Answer 19

Awesome. Thanks a lot. I'll have to vote that coding one as the best response. Lol. How'd you figure that'd even help?

Answer 20

Because... I don't know, a moment of divine intuition? :D

Answer 21

Keep it up. That explained it all for me easily. I program often and that just helped me understand it so easily. Lol.

Answer 22

Although of course, I'm at a loss at how one could code an infinite sum... that coding bit was just an analogy :D

Answer 23

Well, you can continue the while loop and to save CPU Usage you can halt it by about 2 seconds with sleep() or whatever is used for your language for each calculation and go to sleep with your computer on while it does it for you. You'll never reach infinity but each day you'll be satisfied by seeing progress. Either that, or just say the answer is infinity. Lol.

Answer 24

Haha :D
Not all infinite sums are infinity, by the way. For instance
\[\huge \sum_{n=0}^{\infty}\frac1{2^n}\]

Answer 25

If you let this be S, and you, "run this program", it'll be something like
\[\huge S=1+\frac12+\frac14+\frac18+\frac1{16}+......\]

Answer 26

So, if we double S, we get...
\[\huge 2S=2+1+\frac12+\frac14+\frac18+\frac1{16}......\]
And we bring the first 2 to the left side...
\[\huge 2S-2=1+\frac12+\frac14+\frac18+\frac1{16}+...\]Now I want you to observe that the right side of this equation has reverted to the original S. Check it above... we started with
\[\huge S=1+\frac12+\frac14+\frac18+\frac1{16}+......\]So we can now write the right-side of the equation
\[\huge 2S-2=1+\frac12+\frac14+\frac18+\frac1{16}+...\] as...

Answer 27

\[\huge 2S-2=S\]And a simple algebraic way of solving for s yields...
\[\Huge S=2\]So...
\[\Huge S = \sum_{n=0}^{\infty}\frac1{2^n}=2\ne\infty\]

That was fun... good hunting :)

Answer 28

Oh, so they'll converge at 0. Lol. Then make it so that if the number is less that the previous number and is a double then set it to zero.

Answer 29

Converge at 0? No. Converge TO 0, yes... as long as it's the sequence you're talking about :D
But the series (the sum of the terms of the sequence) converges to 2 :)

Answer 30

Anyway, I need to get going now :)
Have fun with Maths ^.^
-----------------------------------
Terence out

Answer 31

Thanks. Ciao. I just finish Math and English. Lol.
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Kris out