Series Concept (1 Viewer)

[xwrathbringerx]

Hey guys

I'm trying to teach myself about the topic series but can't seem to understand it...

Esp. these questions:


1. (x+1)^3 – x^3 = 3x^2 + 3x^ + 1
∑n^2 =

2. (x+1)^4 – x^4 = 4x^3 + 6x^2 + 4x + 1
∑n^3=


Please, culd u please explain how to do these q.s to me??

Thanx a lot!<!-- google_ad_section_end -->

 

[Brontecat]

i second the need for an explanation

although it would be extra helpful if someone could dumb it down and write out each little step

 

[scardizzle]

Hey guys

I'm trying to teach myself about the topic series but can't seem to understand it...

Esp. these questions:


1. (x+1)^3 – x^3 = 3x^2 + 3x^ + 1
∑n^2 =

2. (x+1)^4 – x^4 = 4x^3 + 6x^2 + 4x + 1
∑n^3=


Please, culd u please explain how to do these q.s to me??

Thanx a lot!<!-- google_ad_section_end -->

I would try and help but i have no idea what the question is

what does the 1st expression have to do with the sigma notation also why is

there a power next to the sigma n ?

I've never seen this kind of notation before
<==== my brain when I saw the question

 

[xwrathbringerx]

ummm

For question 1, some of my working out is:

(x+1)^3 - x^3 = 3x^2 + 3x + 1

x = 1 2^3 - 1^3 = 3 * 1^2 + 3 * 1 + 1
x = 2 3^3 - 2^3 = 3* 2^2 + 3 * 2 + 1


x = n (n+1)^3 - n^3 = 3 * n^3 - n^3 = 3 * n^2 + 3 * n + 1

(n + 1 - 1)((n+1)^2 + (n+1) + 1^2) = 3En^3 + (3*n(n+1))/2 + n


-------------------------

Idno ...somehow we're meant to get En^2 = n/6(n+1)(2n+1) but I don't really noe the rite method to get to it

 

[yibbon]

ummm

Idno ...somehow we're meant to get En^2 = n/6(n+1)(2n+1) but I don't really noe the rite method to get to it

Man you need to post the actual question, I have no idea what you need. I can prove that using Induction which is related to series in the course, otherwise I have no clue what you are trying to do.

 

[xwrathbringerx]

lol that was the question:

1. (x+1)^3 – x^3 = 3x^2 + 3x^ + 1


Find ∑n^2.

2. (x+1)^4 – x^4 = 4x^3 + 6x^2 + 4x + 1

Find ∑n^3.

 

[kurt.physics]

lol that was the question:

1. (x+1)^3 – x^3 = 3x^2 + 3x^ + 1


Find ∑n^2.

2. (x+1)^4 – x^4 = 4x^3 + 6x^2 + 4x + 1

Find ∑n^3.

It could mean find the sum of the n^2 terms (although they are in terms of x)?

In this case 1) would be the coefficient of x^2, so the answer is 3.

But that question just seams faulty. Could you say where you got these questions and under what exercise they were under (ie Binomial Coefficients or something else?)

 

[azureus88]

for the first one:

[maths](n+1)^3-n^3=3n^2+3n+1\\\sum_{n=1}^{N}\left [(n+1)^3-n^3 \right ]=\sum_{n=1}^{N}\left [3n^2+3n+1 \right ]\\\sum_{n=2}^{N+1}n^3-\sum_{n=1}^{N}n^3=3\sum_{n=1}^{N}n^2+3\sum_{n=1}^{N}n+N\\(N+1)^3-1=3\sum_{n=1}^{N}n^2+\frac{3}{2}N(N+1)+N\\\sum_{n=1}^{N}n^2=\frac{N^3+3N^2+3N-\frac{3N(N+1)}{2}-N}{3}\\=\frac{N}{6}(N+1)(2N+1)[/maths]

second one should use a similar method.

 

[scardizzle]

for the first one:

[maths](n+1)^3-n^3=3n^2+3n+1\\\sum_{n=1}^{N}\left [(n+1)^3-n^3 \right ]=\sum_{n=1}^{N}\left [3n^2+3n+1 \right ]\\\sum_{n=2}^{N+1}n^3-\sum_{n=1}^{N}n^3=3\sum_{n=1}^{N}n^2+3\sum_{n=1}^{N}n+N\\(N+1)^3-1=3\sum_{n=1}^{N}n^2+\frac{3}{2}N(N+1)+N\\\sum_{n=1}^{N}n^2=\frac{N^3+3N^2+3N-\frac{3N(N+1)}{2}-N}{3}\\=\frac{N}{6}(N+1)(2N+1)[/maths]

second one should use a similar method.

Whoa. I've never seen a question like this before, is this harder 3U?

 

[Trebla]

It uses the idea of a collpasing sum.
Basically all you do is sum both sides of the equation from x = 1 to x = N. The LHS collapses to a simple expression and the RHS can be evaluated using the sum of squares formula and the arithmetic series formula.

 

[xwrathbringerx]

hmmm culd u please show me how to do the 2nd q. too? (so i can verify ive got the right idea going)

Sorry for the trouble

 

[azureus88]

[maths](n+1)^4-n^4=4n^3+6n^2+4n+1\\\sum_{n=1}^{N}\left [ (n+1)^4-n^4 \right ]=\sum_{n=1}^{N}\left [ 4n^3+6n^2+4n+1 \right ]\\\sum_{n=2}^{N+1}n^4-\sum_{n=1}^{N}n^4=4\sum_{n=1}^{N}n^3+6\sum_{n=1}^{N}n^2+4\sum_{n=1}^{N}n+N\\(N+1)^4-1=4\sum_{n=1}^{N}n^3+6\left [ \frac{N}{6}(n+1)(2N+1) \right ]+4\left [ \frac{N}{2}(N+1) \right ]+N\\4\sum_{n=1}^{N}n^3=(N+1)^4-1-6\left [ \frac{N}{6}(n+1)(2N+1) \right ]-4\left [ \frac{N}{2}(N+1) \right ]-N[/maths]

Just expand the terms and you should get an answer