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
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
∑(r3−36r)=∑r3−36∑r
If you know the standard ∑r result then the working should be clear.
For line 3, they took out a common factor of 41n(n+1). Try it and post what you get.
@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.
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?
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) so take that out to give:
n(n+1)[...]
What goes inside the brackets? They have also taken out a factor of 1/4 but try the above first.
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
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