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Tutor profile: Jeffrey O.

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Jeffrey O.
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Subject:Pre-Calculus

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Question:

Find the sum of the first 12 terms of the geometric sequence made by starting with 3 and halving each time (3, 3/2, 3/4, 3/8, ...)

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Jeffrey O.

We can describe this geometric sequence as $$\sum_{k=0}^{n-1} ar^{k}$$ with a=3, r=$$\frac{1}{2}$$, and n=12. Using the formula $$\sum_{k=0}^{n-1} ar^{k}$$=a($$\frac{1-r^{n}}{1-r})$$, we plug in our values for a, r, and n to get that $$\sum_{k=0}^{11} 3(\frac{1}{2})^{k}$$=3($$\frac{1-\frac{1}{2}^{12}}{1-\frac{1}{2}})$$=3($$\frac{1-\frac{1}{4096}}{\frac{1}{2}})$$=3(2)($$\frac{4095}{4096}$$)=5+$$\frac{4090}{4096}$$=5+$$\frac{2045}{2048}$$.

Subject:Discrete Math

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Question:

Prove that if x is an even integer, that x^2 is an even integer as well, and if y is an odd integer, then y^2 is also an odd integer.

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Jeffrey O.

Suppose x is even. By definition, this means that for all integers x, there exists some integer k such that x=2*k. If we square both sides of the equation, we get that x^2=4*k^2. This can be rewritten as x^2=2(2*k^2). Since the integers are closed by multiplication and we know that k is an integer, we know that 2*k^2 is also an integer, since k is an integer. Since we have x^2 written as 2 multiplied by some integer, this fulfills the definition of even, so we know that x^2 must also be even. Then suppose y is odd. By definition, this means that for all integers y, there exists some integer k' such that x=2k'+1. If we square both sides of the equation, we get that y^2=4*k'^2+4*k'+1. This can be rewritten as x^2=2(2*k'^2+2*k')+1. Since the integers are closed by multiplication and addition, and we know that k' is an integer, we know that 2*k'^2+2*k' is also an integer, since k is an integer. Since we have y^2 written as 2 multiplied by some integer +1, this fulfills the definition of odd, so we know that y^2 must also be odd.

Subject:Calculus

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Question:

Find the integral of (x^2+2x-1)/(x-1)*(x^2+1) in terms of x.

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Jeffrey O.

First the fraction has to be split up by partial fraction decomposition. This can be achieved by setting the whole fraction equal to A/(x-1)+(Bx+C)/(x^2+1). Multiplying the right side to have a common denominator, which we can then cancel, gives that (1)x^2+(2)x+(-1)=(A+B)x^2+(C-B)x+(A-C). Since the coefficients for the different powers of x have to be corresponding, we have the equations 1=A+B, 2=C-B and -1=A-C. By solving for these equations, we find that A=1, B=0 and C=2, so we can change our initial fraction (x^2+2x-1)/(x-1)*(x^2+1) into 1/(x-1)+2/(x^2+1). This will allow us to split the original integral into 2 integrals that can be added together, since there is now an addition of 2 separate terms in the integral. The first integral, 1/(x-1), can be done with a u-substitution, setting u=x-1 and du=dx, then integrating 1/u in terms of u to become ln(u), and we can substitute back to get ln(x-1). For the second integral, the 2 can be pulled in front of the integral, and 1/(x^2+1) is a standard integral known to be arctan(x), so if we add the results together along with the constant, we get that the integral is ln(x-1)+2*arctan(x)+c

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