Official TSR Mathematical Society

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Reply 1900

Not to spam this thread but searching for the 2009 STEP III one bore no fuit / Engle (in his problem solving book) states

\displaystyle \sum_{i=1}^{\infty} \frac{1}{F_i} = 4 - \left(\frac{1+\sqrt{5}}{2}\right)

- just added for completeness

Reply 1901

Oh I Really Don't Care

wait - thats not right - ill get back to you when I am in a state that allows me to read pages.

Reply 1902

Elongar

The reciprocal Fibonacci constant has no closed form - I've found that mistake in Engel too...

Reply 1903

DFranklin

Would have been better if he'd said

\sum \frac{1}{F_n} = 5 - \frac{1+\sqrt{5}}{2}

; that would at least be close enough to be convincing (3.36... v.s. 3.38...)

Reply 1904

Elongar

It might be interesting to see whether there is some plausible inductive step (the question was posed in the chapter on induction) that explains its presence in the book...

Reply 1905

KAISER_MOLE

Md. Qela Genisa.

This is a mnemonic for a maths thingy, any guesses as for what? (Hint: Think hot!)

Reply 1906

KAISER_MOLE

Massive hint 'MAGMA'

Just think it's quite a cool word for an algebraic structure. The rest of the mnemonic tells about what conditions you need to include/exclude to move into another structure (Magma-->Quasigroup-->Loop-->Group-->Monoid-->Semigroup-->Magma)

(N=Monoid...which is a bit silly now I look at it)

Reply 1907

Dadeyemi

Here's a simple extension I found to a BMO2 problem.

Suppose we have a circle that passes through each side of a regular n-gon twice. Show that the sum of the lengths of the side of the n-gon to the right of the chords formed where a side of the n-gon intersects the circle is equal to the sum of the lengths to the left of these chords.

Reply 1908

The Muon

A little one which isn't massively mathematical but it quite nice.

You travel from A to B at a speed of 30m/s and then back at a different speed. What speed should you travel back if you wish the average speed of the whole journey to be 60m/s?

I heard it in the pub the other night and so I hope that I got the numbers right :tongue:

Reply 1909

DFranklin

The Muon

Fast. Really fast.

Reply 1910

Simplicity

Isn't that really easy to solve?

(2,3) and (1,1)

Spoiler

This isn't particularly ingenious, but here's an observation I made recently:

Find the minimum value of the constant

\lambda

such that

\displaystyle (a_1 + a_2 + \dots + a_k)^n \le \lambda(a_1^n + a_2^n + \dots + a_k^n)

for

a_i

real and positive.

Hence, write down the value of

\displaystyle \sum_{r_1,r_2,\dots,r_k} \binom{n}{r_1,r_2,\dots,r_k}

where the summation is over all non-negative sequences of integers

r_i, 1 \le i \le k

such that

r_1 + r_2 + \dots + r_k = n

Reply 1913

Elongar

Yup

Spoiler

Reply 1914

Elongar

Yup again

No huge insight required here, but I thought it was quite neat :smile:

Reply 1915

Elongar

Lesson learned....don't wildly copy and paste LaTeX....

Reply 1916

Elongar

you misunderstand :biggrin:

In my haste to crunch through the LaTeX, I copied the summation from the previous part of the reasoning, before attempting to rewrite the whole thing as a binomial expansion. So in essence, due to a general lack of brain function, for a short while I thought that each term in the summation could be evaluated as a binomial expansion (which is obviously not true). I didn't spot it until right now though.

Reply 1917

rnd

What's that 5^k doing? I don't think it's needed.

What you have is half of a row of Pascal's triangle. Easy to see when n is odd. and since the full row is divisible by 2^n etc....

What I haven't thought about yet is showing that your sum is half the row total when n is even.

What's that spoiler about by the way?