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Class of 2025 (2025 HSC CHAT) (28 Viewers)

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
 

mookie9882882

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guys i need good pen recs my handwriting is shit and i need a pen thats not too inky but also really good
 

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