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# Help with MaTh

-1
1488
3
+87

1. Find all $$r$$ for which the infinite geometric series $$2 + 6r + 18r^2 + 54r^3 + \dotsb$$
is defined. Enter all possible values of $$r$$ as an interval.

2. Compute $$1 - 2 + 3 - 4 + \dots + 2005 - 2006 + 2007$$

3. Let $$f(x) = \frac{x^2}{x^2 - 1}$$ Find the largest integer $$n$$ so that $$f(2) \cdot f(3) \cdot f(4) \cdots f(n-1) \cdot f(n) < 1.98$$

Dec 6, 2019

#1
0

1. You want the geometric series to converge, so -1 < r < 1.

2. 1 - 2 + 3 - 4 + ... + 2005 - 2006 + 2007 = (1 - 2) + (3 - 4) + ... + (2005 - 2006) + 2007 = 1002.

3. The product will telescope, and you are left with n/(n - 1), so you want n/(n - 1) < 1.98.  The largest n that satisfies this is n = 201.

Dec 6, 2019
#2
+8460
+1

1)
Common ratio = $$\dfrac{6r}{2}$$ = 3r

-1 < (Common ratio) < 1

-1 < 3r < 1

$$\dfrac{-1}{3} < r < \dfrac{1}{3}$$

Dec 6, 2019
#3
+8460
+1

2)

Consider the pairwise sum of terms.

$$\quad 1-2+3-4+\cdots+2005-2006+2007\\ =(1-2)+(3-4)+\cdots+(2005-2006)+2007\\ =-1+(-1)+\cdots+(-1)+2007\\ \text{There are in total }\dfrac{2006}{2}=1003\text{ pairs that sums to }-1.\\ = -1(1003) + 2007\\ = 1004$$

Dec 6, 2019