#1
Hi guys, i've only come here as a last resort so if anyone could help me i'd be extremely grateful

This is Futher maths core 1:
I'll use 'Z' in the place of sigma and all the brackets are just to ensure there isn't any confusion over squared numbers etc

I have a series question that asks to find from 1-n the Z of (r^3 - r^2)

Here are the basic formulas for each-

r^3 = (1/4)(n^2)[(n+1)^2]

r^2= (1/6)(n)(n+1)(2n+1)

And the form that it has to be in is-

(1/12)(n)([n^2] -1)(3n+2)

From the solutions I am told that both (n) and (n+1) are roots

So what I have is this-

Step 1- ( (1/4)(n^2)[(n+1)^2]) - [(1/6)(n)(n+1)(2n+1)]

Step 2- I can take out 1/12 and n as a common factor so we're left with...

(1/12)(n) [(3)(n)[(n+1)^2]) - [(2)(n+1)(2n+1)]]

However from there I'm really lost as to where to go with it; so if anyone can point out what i need to do to get the answer (if the format that i've laid it out in makes sense) , i'd hugely appreciate it

#4
Distribute all the coefficients and FOIL and whatnot, then factor what you're left with?
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#5
I finally figured it out, i'd been a tool and forgotten about the difference of two squares; here's the solution:

( (1/4)(n^2)[(n+1)^2]) - [(1/6)(n)(n+1)(2n+1)]

Take out 1/12 and n


(1/12)(n) [(3)(n)[(n+1)^2]) - [(2)(n+1)(2n+1)]]

Then take out n+1


(1/12)(n)(n+1) [(3)(n)[(n+1)]) - [(2)(2n+1)]]

Sort the brackets out


[(3)(n)[(n+1)]) - [(2)(2n+1)]]
=3n^2 + 3n -4n -2
=3n^2 -n -2
=(3n+2)(n-1)

Put that back in so you get

(1/12)(n)(n+1) [(3n+2)(n-1)]

(n+1)(n-1) = (n^2 -1)

So (1/12)(n) [(3n+2)(n^2 -1)] as required
#7
are you doing Maths at uni?
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