Triangle and Limit

kunny

Given sequence of points $A_n\ (a_n,\ a_n^2)\ (n = 1,\ 2,\ \cdots)$ and $B_n\ (b_n,\ 0)\ (n = 0,\ 1,\ 2,\ \cdots)$ . Suppose that $A_n$ are on the parabola $y = x^2$ and approach to the origin from right hand; $B_n$ are on the $x$ axis and approach in such way as well. If $\triangle{A_nB_nB_{n - 1}}$ form an equilateral triangle for every $n = 1,\ 2,\ \cdots$ and $a_1=1$ , then
Find
$\sum_{n = 1}^{\infty} a_n^2\ \text{and}\ \sum_{n = 1}^{\infty} a_n^3.$

**Posted:** Tue Nov 03, 2009 9:51 am

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Today's calculation of Integral Digest

kenn4000

$(a_n-b_n)\sqrt 3=a_n^2=(b_{n-1}-a_n)\sqrt 3$

$b_{n-1}-b_n=\frac{2a_n^2}{\sqrt 3}$

so $\sum_{n=1}^N a_n^2= \frac{\sqrt 3}{2}\sum_{n=1}^N (b_{n-1}-b_n)=\frac{\sqrt 3}{2}(b_0-b_{N-1}) \to \frac{\sqrt 3}{2}b_0=\frac...$

i dont see the cube part though =/ something to do with the area?

**Posted:** Tue Nov 03, 2009 11:36 pm

AndrewTom

Mid-point of base of triangle n is $\frac{b_{n} + b_{n-1}}{2}$

$a_{n} ^{2} = (\frac{b_{n} + b_{n-1}}{2})^{2}$

$a_{n} = \frac{b_{n} + b_{n-1}}{2}$

But $b_{n-1} - b_{n} = \frac{2a_{n}^{2}}{\sqrt{3}}$ (from kenn4000's post)

Therefore $b_{n-1} - b_{n})(b_{n-1} + b_{n}) = \frac{4a_{n}^3}{\sqrt{3}}$

$\frac{4}{\sqrt{3}} \sum_{n=1}^{\infty} a_{n}^{3} = \sum_{n=1}^{\infty} b_{n-1}^{2} - b_{n}^{2}$

$=b_{0}^{2} = (1+ \frac{1}{\sqrt{3}})^{2}$

$= \frac{4+2\sqrt{3}}{3}$

Therefore $\sum_{n=1}^{\infty} a_{n}^{3} = \frac{2\sqrt{3} + 3}{6}$ .

**Posted:** Wed Nov 04, 2009 5:14 am