<p>The first term of the sequence is 3/4, the second term is 3/2
Find out the 17th term</p>
<p>Find the Sum of the first 15 terms</p>
<p>24, 8 , 8/3, 8/9
Find the first term that is less then 0.0001- dont just multiply by 3 to get it</p>
<p>Explain why X^2+6x+7 is still an inverse function from -3, - infinity</p>
<p>and the final one which is the hardest one</p>
<p>You have a series : m m+3 m+6 ... 4m, prove why 5x is greater then 1, 2, 3...m</p>
<p>help on any of these problems would be greatly appreciated</p>
<p>Most of these problems are ill-posed, but I can guess what you meant for a couple of them.
For the sequence 3/4, 3/2, ... you haven't stated whether it's geometric, arithmetic, or neither, but I'll guess geometric.
For the geometric series a + ar + ar^2 + .... if you count a as the first term (in some formulas it's the zeroeth)
the nth term is ar^(n-1) and the nth partial sum is
Sn = a(r^n - 1)/(r-1)
For your problem, a = 3/4 and r = 2, so the 17th term is (3/4)*2^16,
and the sum of the first 15 terms is
S15 = (3/4)(2^15 - 1) .</p>
<p>The sequence 24, 8, 8/3, 8/9 ... is pretty clearly supposed to be geometric with a = 24 and r = 1/3. The nth term is
24*(1/3)^(n-1) and you want this less than .0001.
Take the log of both sides (either base 10 or natural); log is a monotone function so it preserves the inequality:
(n-1)log(1/3) < log (.0001/24)
Divide both sides by log 1/3, but be careful to change the sign of the inequality because log 1/3 is negative:
n-1 > [log(.0001/24)]/log(1/3)]
then add one to both sides.</p>
<p>On the interval (-infinity, -3) the function x^2 + 6x + 7 could be considered an invertible function (if you say it is undefined on the other half of its normal domain) because it passes the horizontal line test/is monotone on that interval.
You can see this because the vertex is at (-3, -2). The parabola goes down to the left of -3 and goes up to the right.</p>
<p>I have no idea what you mean in the last problem. What is x? Obviously you'll be using the formulas for an arithmetic series when you pose the question correctly.</p>