# HSC 2017 MX1 Marathon (1 Viewer)

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#### pikachu975

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

##### Well-Known Member
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

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

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

#### 1729

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

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

#### pikachu975

##### I love trials
Moderator
here's a pree cool nested root integral to try out;

$\bg_white \noindent Evaluate \int_{2}^{12}\sqrt{x+\sqrt{x+\sqrt{x+...}}} dx$
Woahh you do nested root integrals in general maths cough

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#### shehan123

##### Member
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?

#### 1729

##### Active Member
$\bg_white \sqrt{1+5\sqrt{1+6\sqrt{1+7\sqrt{1+\ldots}}}} = \ ?$

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