The upper limit can be written as y = 10, though the "y" is usually dropped for the upper limit.
With y - 2 = k what you do is ask what is the value of k when y = 3 (k = 3 - 2 = 1), and what is the value of k when y = 10 (k = 10 - 2 = 8).
\(\sum_{k=2}^{k=9}\frac{1}{k}=\sum_{k=1}^{k=9}\frac{1}{k}-\frac{1}{1} =\sum_{k=1}^{k=8}\frac{1}{k}-\frac{1}{1}+\frac{1}{9}\)
It's perhaps a little more elegant the following way:
\(\sqrt{36x^{36}}=\sqrt{6^2x^{18\times2}}=\sqrt{6^2(x^{18})^2}=6x^{18}\)
I think it means you just keep putting the number you calculate back into the function until the result is 1. i.e. C(9) = 28, C(28) = 14, C(14) = 7, ...etc.
Here is a short computer program that does the calculations:
Like so:
Do you mean like this? (A matrix approach seems like overkill here!)
You have the following two equations:
s = c (s=nbr of sausage buns, c = nbr of custard buns)
c + 27 = 6(s - 68)
Can you take it from here?
@wisdom321: Note that your equation, y^2 + 9y - 18 = 0, shoud be y^2 + 9y + 18 = 0
I set j = k-2, so k=3 becomes j = 1, and k = n becomes j = n-2. Then:
The sum \(\Sigma_{j=1}^{m}j\) is given by \(m(m+1)/2\), so when m = n-2, the sum is (n-2)(n-1)/2.
As 23 is a prime number we must have:
\((23x+a)(x+b)=23x^2+kx-5\)
Multiply out the left-hand side
\(23x^2+(a+23b)x+ab=23x^2+kx-5\)
Compare coefficients on both sides.
Can you take it from here, remembering that the coefficients must be integers.
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