HSC 2017 MX1 Marathon (2 Viewers)

pikachu975 · Oct 1, 2017

kawaiipotato said:
$\noindent Hint: Expand using the Binomial Theorem to get $(2x^3 - \frac{1}{x})^n \equiv \sum_{k=0}^n \binom{n}k 2^{k}(-1)^{n-k}x^{4k -n}$. Consider $x^j$ for $j\geq 0$ where $j$ is an expression of $k,n$. When $j=0$, what does this say about $k$ and $n$?$

Where'd you get x^(4k-n)? The general term would have (2x^3)^(n-k) (1/x^k) which would be 3n - 4k wouldn't it

InteGrand · Oct 1, 2017

pikachu975 said:
Where'd you get x^(4k-n)? The general term would have (2x^3)^(n-k) (1/x^k) which would be 3n - 4k wouldn't it

He put the "k" with the x^3 part instead of with the 1/x part.

(Remember, you can put it with either part.)

jathu123 · Oct 5, 2017

here's a pree cool nested root integral to try out;

$\noindent $ Evaluate $\int_{2}^{12}\sqrt{x+\sqrt{x+\sqrt{x+...}}}$ d$x$

1729 · Oct 5, 2017

jathu123 said:
here's a pree cool nested root integral to try out;

$\noindent $ Evaluate $\int_{2}^{12}\sqrt{x+\sqrt{x+\sqrt{x+...}}}$ d$x$

$$\noindent Let $u = \sqrt{x+\sqrt{x+\sqrt{x+...}}} \implies u^2 - u - x = 0 \implies u = \frac{1+\sqrt{1+4x}}{2}$, taking only the positive answer since $u$ is a positive square root. \\\\ $\int_{2}^{12}\sqrt{x+\sqrt{x+\sqrt{x+...}}}$ d$x = \frac{1}{2} \int_{2}^{12} 1+ \sqrt{1+4x}$ d$x = \frac{1}{2}\cdot \frac{188}{3} = \frac{94}{3}$

pikachu975 · Oct 5, 2017

jathu123 said:
here's a pree cool nested root integral to try out;

$\noindent $ Evaluate $\int_{2}^{12}\sqrt{x+\sqrt{x+\sqrt{x+...}}}$ d$x$

Woahh you do nested root integrals in general maths

cough

shehan123 · Oct 5, 2017

pikachu975 said:
Where'd you get x^(4k-n)? The general term would have (2x^3)^(n-k) (1/x^k) which would be 3n - 4k wouldn't it

Was the answer to this, multiples of 4?

jathu123 · Oct 5, 2017

shehan123 said:
Was the answer to this, multiples of 4?

Yes

1729 · Oct 8, 2017

dan964 · Oct 10, 2017

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HSC 2017 MX1 Marathon (2 Viewers)

pikachu975

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