Let x1, ..., x25 be such positive integers that x1⋅x2⋅ ... ⋅x25 = x1 + x2 + ... + x25. What is the maximum possible value of the largest of numbers x1, x2, ..., x25?
Solution to the problem tells us that ##S_5 + i S_6## is the sum of the terms of a geometric sequence and thus the solutions should be :
$$S_5 = \frac{\sin( (n+1) x)}{\cos^n(x) \sin(x)},\,\,\,\, S_6 = \frac{\cos^{n+1}(x) - \cos((n+1)x)}{\cos^n(x) \sin(x)} , x \notin \frac{\pi}{2} \mathbb{Z}$$...
I'm using the sum of a geometric series formula, but I'm not sure how to find the ratio, r. The n is confusing me.
The solution is below, but I'm having trouble with the penultimate step.
The matrix ##A## in question is
##\dfrac{1}{3}
\left(\begin{array}{rrr}
-2 & 1 & -2 \\
-2 & -2 & 1 \\
1 & -2 & -2
\end{array}\right)##
One can easily verify that ##AA^t=I##, hence an isometry. To find its geometric meaning, one can proceed to find ##U=\text{ker} \ (F-I)=\text{ker} \...
My question is Why is the sum to infinity used as opposed to Sum to n? and How can I deduce that the sum to infinity must be used from the question?Total Distance = h + 2*Sum of Geometric progression (to infinity)
h + 2*h/3 / 1-1/3
h + 2h/3 *3/2 = h + h = 2h
At first I did sum to infinity...
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And AP'=BP'=BD/sin(a) and BP=BD/sin(a+b) and AP=BD/sin(a-b).
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After doing some computations, I stuck here. $\frac{2(\log a+\log r)}{\log a+2\log r}=\frac{2(\log a)^2+3\log r\log a +2(\log r)^2}{(\log a)^2+\log r\log a}$
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(x+2/x)=(x+3)/(x+2)
(x+2)2=x2+3x
x2+2x+4=x2+3x
x=4
so i think r=(x+2)/x
putting x=4
r=3/2
also...
Mentor note: Thread moved from a different forum section, so missing the homework template.
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