Class of 2025 (2025 HSC CHAT) (35 Viewers)

melanie_o

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The offer is only given based off if ur year 11 marks were good enough for the course (you will be given the prep school if u dont qualify for any course). my mate got rejected from law (she got Cs and Bs at a Non gateway school) but instead got into criminology. This is more targeted towards students who have fewer opportunities, lack of funding and support. The fact u compared the average atar of 70 to a mystery mark gives it away that u clearly don't understand class/ financial differences and the whole basis of gateway being equity. Your complaint feels like a projection that u fear not getting into ur course due to a "retard" taking ur place. It's honestly a shame people believe UNSW is lowering their standards when they're actually helping disadvantaged people.
This! The reason why Gateway has changed my life is because they provide finantial support. There is no uni near me within 90km, and the lack of teachers at my school has meant that I have had basically self-study most of the year.
 

alphxreturns

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nah like prove 2n c n = so and so
the 'given (1+x)^2n = (1+x)^n (1+x)^n'?

look for coefficients, look for substitutions for x, look for anything thats recognisable from either lhs or rhs. like derivatives or integrals
 

alphxreturns

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

expanding LHS

(2n+1)C0 × X^0 + (2n+1)C1 × X^1 + ... + (2n+1)Cn × X^n + ... + (2n+1)C(2n+1) × X^2n+1

symmetry of Pascal's triangle means nC0 = nCn
and so on, (2n+1)C0 = (2n+1)C(2n+1)

so since 2n+1 is odd, (sub n = any number if u want proof)
letting X = 1,
the expansion is 2((2n+1)C0 + ... + (2n+1)Cn)
rhs = (1+1)(1+1)^2n

so 2n+1C0 +...+ 2n+1Cn = 4^n

sorry I'm pressed on time but the main part was symmetry and X=1
 

ros545

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Hey, is early entry for UTS closing today at 11:59pm (7th September) or tomorrow at 11:59pm (8th September)?
Thanks
 

bigupsanky

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

expanding LHS

(2n+1)C0 × X^0 + (2n+1)C1 × X^1 + ... + (2n+1)Cn × X^n + ... + (2n+1)C(2n+1) × X^2n+1

symmetry of Pascal's triangle means nC0 = nCn
and so on, (2n+1)C0 = (2n+1)C(2n+1)

so since 2n+1 is odd, (sub n = any number if u want proof)
letting X = 1,
the expansion is 2((2n+1)C0 + ... + (2n+1)Cn)
rhs = (1+1)(1+1)^2n

so 2n+1C0 +...+ 2n+1Cn = 4^n

sorry I'm pressed on time but the main part was symmetry and X=1
oh ye u right ty ty
 

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