The Student Room Group

maths help

https://imgur.com/a/CJeGEtn
https://imgur.com/a/G9SrYFR
i have done a, but for b im confused.
i got to making it (r^3 - 36r) , i just expanded it.
and i got to that = the standard summation for r^3 + the summation for 36r.
the mark scheme has the summation for 36r as (-36/2)n(n+1) but i dont understand how you get that from the standard summation of r?
and i also have no idea how they simplified it in line 3, thank you

Scroll to see replies

Original post by Ask Jeeves
did you square root the imperial function?


sorry i have no idea what that means
Reply 2
Original post by Gent2324
https://imgur.com/a/CJeGEtn
https://imgur.com/a/G9SrYFR
i have done a, but for b im confused.
i got to making it (r^3 - 36r) , i just expanded it.
and i got to that = the standard summation for r^3 + the summation for 36r.
the mark scheme has the summation for 36r as (-36/2)n(n+1) but i dont understand how you get that from the standard summation of r?
and i also have no idea how they simplified it in line 3, thank you

(r336r)=r336r\displaystyle \sum \left( r^3 - 36r\right) = \sum r^3 - 36\sum r

If you know the standard r\sum r result then the working should be clear.

For line 3, they took out a common factor of 14n(n+1)\frac{1}{4}n(n+1). Try it and post what you get.
Original post by Gent2324
sorry i have no idea what that means

Square rooting is where you divide a number by itself.
Reply 4
Original post by Gent2324
sorry i have no idea what that means

Troll post.
Original post by Ask Jeeves
Square rooting is where you divide a number by itself.


I was referring to when you said imperial function and why I would do that.

And I’m pretty sure that I would always get 1 with your definition of a square root
Original post by Notnek
Troll post.

Im just trying to help, sorry If it doesn't work :frown:
Original post by Notnek
Troll post.


Oh ok, I’m out atm so the post looks all weird since I’m not on a pc, will give it a go when I get back
Reply 8
@Gent2324 please use better thread titles so that people know whether to click on your thread, plus your question can be found by students in the future. Thanks.
Reply 9
Original post by Gent2324
Oh ok, I’m out atm so the post looks all weird since I’m not on a pc, will give it a go when I get back

Best to not use the app in the maths forum. Use the mobile site instead.
Reply 10
Original post by Notnek
Best to not use the app in the maths forum. Use the mobile site instead.

@Doonesbury do you know if LaTeX works on the old app?
so, if you're adding successive terms of multiples of 3, say
you'd have
3 + 6 + 9 + 12 + 15 + .... + 3n
=3(1+2+3+4+5+...+n)
=3. n(n+1)/2

Does that help you understand why we get 36n(n+1)/2 ?
[it'll be negative because the LHS is SUM(r^3) subtract SUM(36r) ]

Thinking about going from the MS line 2 to line 3:
What factors are common to both terms in line 2?
Clue : one of the common factors is n/4

So, can you completely factorise line 2?
Reply 12
Original post by Notnek
@Doonesbury do you know if LaTeX works on the old app?


No it doesn't. You just get a :smile:
Reply 13
Original post by Doonesbury
No it doesn't. You just get a :smile:

Thanks. In that case I'm not holding out much hope for it to work on the new app in the next few years :frown:
Original post by Notnek
(r336r)=r336r\displaystyle \sum \left( r^3 - 36r\right) = \sum r^3 - 36\sum r

If you know the standard r\sum r result then the working should be clear.

For line 3, they took out a common factor of 14n(n+1)\frac{1}{4}n(n+1). Try it and post what you get.


Original post by begbie68
so, if you're adding successive terms of multiples of 3, say
you'd have
3 + 6 + 9 + 12 + 15 + .... + 3n
=3(1+2+3+4+5+...+n)
=3. n(n+1)/2

Does that help you understand why we get 36n(n+1)/2 ?
[it'll be negative because the LHS is SUM(r^3) subtract SUM(36r) ]

Thinking about going from the MS line 2 to line 3:
What factors are common to both terms in line 2?
Clue : one of the common factors is n/4

So, can you completely factorise line 2?

so with the standard summation of r natural numbers, if i wanted 2r, it is : 2(1/2n(n+1) ?

if thhats the case i see why its 36/2 n .
as for factorising i still dont really see what going on, they see mto take a square out but what do they do with the square afterwards?
Reply 15
Original post by Gent2324
so with the standard summation of r natural numbers, if i wanted 2r, it is : 2(1/2n(n+1) ?

if thhats the case i see why its 36/2 n .
as for factorising i still dont really see what going on, they see mto take a square out but what do they do with the square afterwards?

It's basic factorising but disguised as something harder:

Unparseable latex formula:

\dipslaystyle \frac{1}{4} \times n \times n \times (n+1) \times (n+1) - \frac{36}{2}\times n\times (n+1)



There is a common factor of n(n+1)n(n+1) so take that out to give:

n(n+1)[    ...    ]\displaystyle n(n+1)\left[ \ \ \ \ ... \ \ \ \ \right ]

What goes inside the brackets? They have also taken out a factor of 1/4 but try the above first.
Original post by Notnek
It's basic factorising but disguised as something harder:

Unparseable latex formula:

\dipslaystyle \frac{1}{4} \times n \times n \times (n+1) \times (n+1) - \frac{36}{2}\times n\times (n+1)



There is a common factor of n(n+1)n(n+1) so take that out to give:

n(n+1)[    ...    ]\displaystyle n(n+1)\left[ \ \ \ \ ... \ \ \ \ \right ]

What goes inside the brackets? They have also taken out a factor of 1/4 but try the above first.

https://imgur.com/a/NmPjh6P
Reply 17

Yes that's right and you're very close to what they have. Next take 1/4 outside the brackets. A quick way to do that is to multiply each term by 4.
Original post by Notnek
Yes that's right and you're very close to what they have. Next take 1/4 outside the brackets. A quick way to do that is to multiply each term by 4.


ah yes i got it now thank you, im also stuck on the next question unfortunately, https://imgur.com/a/XUNAmUL
im on part C and for some reason they integrate y-1 but i have no idea why. i thought that since K = 4 , you would integrate pi(4x^2+1)^2 with limits of 1.16 and 1 since thats the height of the bowl? i have no clue as to where they got y-1 from in the answer
https://imgur.com/a/4Rm9g1k
Reply 19
Original post by Gent2324
ah yes i got it now thank you, im also stuck on the next question unfortunately, https://imgur.com/a/XUNAmUL
im on part C and for some reason they integrate y-1 but i have no idea why. i thought that since K = 4 , you would integrate pi(4x^2+1)^2 with limits of 1.16 and 1 since thats the height of the bowl? i have no clue as to where they got y-1 from in the answer
https://imgur.com/a/4Rm9g1k

Volume of revolution about the y-axis:

πx2 dy\displaystyle \pi \int x^2 \ dy

Quick Reply

Latest