I've just received my AS level results and they were good generally. However I was very disappointed in my Maths A level results seeing as my overall grade was a C, but I got an A in C1, an A in C2 but a U in S1 I've decided that my mathematical capability in core is strong enough to get me through both AS and A2 further maths in one year but just wanted to know what people think. Thanks
I've just received my AS level results and they were good generally. However I was very disappointed in my Maths A level results seeing as my overall grade was a C, but I got an A in C1, an A in C2 but a U in S1 I've decided that my mathematical capability in core is strong enough to get me through both AS and A2 further maths in one year but just wanted to know what people think. Thanks
I think you should concentrate in the Straight Maths
I made a thread similar to this so I'll summarise. Yes it is doable, that being said it is not encouraged as it is very difficult to do (you will be doing 9 Maths exams next year). Take it from someone in your position, carry your AS subjects onto A2, work harder than you did last year and get that C up to an A. Another point, modules like FP2/3 and M4/5 are an order of magnitude harder than C1/2 so they will take a significant amount of time to get up to standard. I you need any advice just PM me P.S. I'm doing FM in a year (with A2 Maths and A2 Physics)
dammit, that idea crossed my mind i just didn't see it through! thank you!
hope you dont mind helping me out on this question:
2) Given that for all real values of r, (2r+1)^3 - (2r-1)^3 = Ar^2 + B, where A and B are constants. Find the value of A and B. I did the whole r=1 etc and it did work up until I was left with (2n+1)^3 - 1 and even after expanding I was left with a cubic which doesn't match up to Ar^2 + B.
dammit, that idea crossed my mind i just didn't see it through! thank you!
hope you dont mind helping me out on this question:
2) Given that for all real values of r, (2r+1)^3 - (2r-1)^3 = Ar^2 + B, where A and B are constants. Find the value of A and B. I did the whole r=1 etc and it did work up until I was left with (2n+1)^3 - 1 and even after expanding I was left with a cubic which doesn't match up to Ar^2 + B.
expand the LHS and compare coefficients
or
set r =1/2 and then set r= -1/2
I am not answering more as I am watching and murder mystery film and I need to follow the plot. goodnight
hope you dont mind helping me out on this question:
2) Given that for all real values of r, (2r+1)^3 - (2r-1)^3 = Ar^2 + B, where A and B are constants. Find the value of A and B. I did the whole r=1 etc and it did work up until I was left with (2n+1)^3 - 1 and even after expanding I was left with a cubic which doesn't match up to Ar^2 + B.
Since this is supposed to be an identity and you only have 2 constants to find, you can plug in any 2 values of r e.g. r = 0 and r = 1 and then solve for A and B.
Alternatively (but longer) just expand the 2 brackets on the LHS and work out A and B explicitly. Note that you should certainly NOT end up with a cubic - the 1st term in the 1st bracket will be 8r^3 and the 1st term in the 2nd bracket will come out as 8r^3 too, so they will cancel when you take one from the other