A nice 4U Q16 (1 Viewer)

[integral95]

Nope cause the second expression if you expanded properly is x(1+2x+3x^2+…(N-1)x^(N-2))

Which is not a gp but... You know

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Yeah i just saw the solution and i get what you mean

I did it like this lol..... after expanding everything you get



And simplified everything from there

 

[aDimitri]

Yeah i just saw the solution and i get what you mean

I did it like this lol..... after expanding everything you get



And simplified everything from there

yeah uh i just summed the GP twice.
this is a nice solution though

 

[dunjaaa]

Same solution

 

[emilios]

I can't do part ii wtf. Dunja how did you go from the 3rd line to the 4th line of working out; how do you simplify 1+2x+3x^2 + ... (n-1)x^(n-2) . I don't see how you can use the result of pt i since the coefficients are different

 

[Fade1233]

I can't do part ii wtf. Dunja how did you go from the 3rd line to the 4th line of working out; how do you simplify 1+2x+3x^2 + ... (n-1)x^(n-2) . I don't see how you can use the result of pt i since the coefficients are different

Wait emilios, remember binomial differentiation, I think you have to differentiate both sides. And then you can use sum of gp as said above for the other part.

 

[mreditor16]

emilios, here are the official solutions for what you are perplexed by

 

[mreditor16]

Wait emilios, remember binomial differentiation, I think you have to differentiate both sides. And then you can use sum of gp as said above for the other part.

idgi. whats all this talk about differentiation?

if you look at the solutions -->

4unitmaths.com/fort-st-ext2-2013.pdf

you will most clearly see that differentiation is not required, from what I can see. and I followed the same procedure as the provided answers.....

EDIT - Unless there's been something that I've missed. :/

 

[emilios]

emilios, here are the official solutions for what you are perplexed by

OK everything looks good except going from the 1st to 2nd line. Are we expanding? Shouldn't the expansion be n + nx -x +nx^2+.... (i.e. dunja's working out). I mean can you explain to me with words what we're actually supposed to do?

 

[mreditor16]

OK everything looks good except going from the 1st to 2nd line. Are we expanding? Shouldn't the expansion be n + nx -x +nx^2+.... (i.e. dunja's working out). I mean can you explain to me with words what we're actually supposed to do?

nup, we're not expanding. just separating out the terms in the integral, in order to use i) and identity that is asked to be assumed at beginning of ii)

do you get what i mean? its hard to explain in words. haha

 

[emilios]

What do you mean 'separating out terms in the integral'? What happens to the n's in the coefficients?

 

[mreditor16]

What do you mean 'separating out terms in the integral'? What happens to the n's in the coefficients?

how bout this. try getting from this to this, as indicated by the arrow. like do gathering of terms:



if that makes sense, now look at the original solution.....

 

[emilios]

how bout this. try getting from this to this, as indicated by the arrow. like do gathering of terms:



if that makes sense, now look at the original solution.....

OHH LOL i get it, but i feel so dumb now. sorry my brain's been switched off recently

 

[dunjaaa]

You know that 1+x+x^2+...+x^n = (1-x^n)/(1-x) {Sum of GP}. Differentiate both sides and you obtain that expression.

 

[dunjaaa]

Note: We can re-write the integral in this manner n+(n-1)x+(n-2)x^2+...+(n-(n-2))x^(n-2)+(n-(n-1))x^(n-1). Expanding, you obtain n(1+x+x^2+...+x^(n-1))-x(1+2x+3x^2+...+(n-1)x^(n-2))

 

[tyrone97]

why are there equality signs for part iii

 

[emilios]

Note: We can re-write the integral in this manner n+(n-1)x+(n-2)x^2+...+(n-(n-2))x^(n-2)+(n-(n-1))x^(n-1). Expanding, you obtain n(1+x+x^2+...+x^(n-1))-x(1+2x+3x^2+...+(n-1)x^(n-2))

Yeah this is probably the most sensible method (e.g. the one I would think of in an exam). Thanks a lot dunja!

 

[Fade1233]

You know that 1+x+x^2+...+x^n = (1-x^n)/(1-x) {Sum of GP}. Differentiate both sides and you obtain that expression.

Like I also said but Mreditor disagreed