Girls vs boys maths challenge

Prove that

$[br]\sum\limits_{r=0}^nr{n\choose r}=n2^{n-1}[br]$
.

Fair enough but you've complicated it more than what was needed

I don't like induction

Consider

$\displaystyle \begin{aligned} f(x) = (x+1)^{n} = \sum_{r=0}^{n} \binom{n}{r} x^{r} \implies f'(1) & = \sum_{r=0}^{n} \binom{n}{r} \cdot r \\ & = n \cdot (x+1)^{n-1} \bigg|_{x = 1} \\ & = n \cdot 2^{n-1} \end{aligned}$

as required.

3987^12 + 4365^12 = ((3987+4365)^2-(2x3987x4365))^6
Rounding both numbers:
We get (8000^2 - 32000000)^6 = 32000000^6 =approx.( 3x10^7)^6 = 729 x 10^42 = 7.3 x 10^44
4472^12 = approx. (5x10^3)^12 = 244142875 x 10^36 = approx 2.5x10^44
Even by constantly rounding up the values for 4472^12 and rounding down 3.2 to 3 for the other expression- which is magnified a lot given the indices involved, 4472^12 < 3987^12 +.4365^12 (the difference between the two is far too big even given the big effect rounding will have had for them to be equal).

ETA: Hah, the difference is about 10^33 :P

Interestingly though, they match in the first 10 decimal places as well as the last decimal place. The accuracy is in the region of $10^{-8}$% It's a 'near-solution'. There was another similar one in the background of the Simpsons episode where homer goes behind a bookshelf and ends up in another dimension (and then ends up in the real world) along with loads of other things like P=NP.

Your initial factorization is incorrect. Also, you have not justified that your rounding did not create the error.

Interestingly though, they match in the first 10 decimal places as well as the last decimal place. The accuracy is in the region of $10^{-8}$% It's a 'near-solution'. There was another similar one in the background of the Simpsons episode where homer goes behind a bookshelf and ends up in another dimension (and then ends up in the real world) along with loads of other things like P=NP.

Fair enough for your second point but a^2 + b^2 = (a+b)^2 - 2ab
a^12 + b^12 = (a^2 + b^2)^6, or not? So a^2+ b^2 may be written as:
( (a+b)^2 - 2ab)^6. No?

No. That requires $(a^2+b^2)^6 = a^{12}+b^{12}$, which implies that (1+1)^6 = 1+1 - clearly rubbish.

Right. I was going to use the binomial expansion but thought that was a shortcut- out of interest, what is (a^2 + b^2)^6?

Binomial expansion:
$a^{12} + 6 a^{10} b^2 + 15 a^8 b^4 + 20 a^6 b^6 + 15 a^4 b^8 + 6 a^2 b^{10} + b^{12}$

Thanks So just the normal expansion- not like (k+3x)^7 or something similar?

I don't like induction

Consider

$\displaystyle \begin{aligned} f(x) = (x+1)^{n} = \sum_{r=0}^{n} \binom{n}{r} x^{r} \implies f'(1) & = \sum_{r=0}^{n} \binom{n}{r} \cdot r \\ & = n \cdot (x+1)^{n-1} \bigg|_{x = 1} \\ & = n \cdot 2^{n-1} \end{aligned}$

as required.

Always like that, $(a+b)^{n} \not\equiv a^{n} + b^{n}$

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Yet (a^b)^c = a^bc. It's so ANNOYING!

Always like that, $(a+b)^{n} \not\equiv a^{n} + b^{n}$

Maths question: A-Level
(should be simple)

A curve, y=f(x) passes through point (0,4) and is such that f'(x) = 8x³ + 4x - 2

Find f(x)