http://i51.tinypic.com/2v94tgn.gif
I used tinypic since they'd be a bit hard to format here.
Anyway, the first one I don't understand at all really. I know the deformed E thing means it's a series, but that's all I can get out of it.
The second one makes more sense, but I don't know what the n is greater or equal to 3 thing means.
Whoever can answer this gets 10 pts.
I used tinypic since they'd be a bit hard to format here.
Anyway, the first one I don't understand at all really. I know the deformed E thing means it's a series, but that's all I can get out of it.
The second one makes more sense, but I don't know what the n is greater or equal to 3 thing means.
Whoever can answer this gets 10 pts.
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You can think of this as a series, but you can ALSO think of it as a function that ONLY takes inputs of positive integers. That might make it a little easier to see.
For the first question, you just plug in k=1, and you get (-1)^2 * 1, which is 1.
Then you plug in k=2, and you get (-1)^3 * 2, which is -2.
Keep plugging in all the k's until you get to k=6.
Then what they're wanting you to do is write the sum out, so you write 1 + -2 + ...
You'll have six terms total in your sum.
One hint: whenever you see (-1)^n or (-1)^(n+1), it acts like flipping a lightswitch. (-1)^n will be -1 for odd n, and will be +1 for even n, so your odd terms will be -, and your even terms will be +. Likewise, (-1)^(n+1) will do the opposite: odd terms positive, even terms negative. (-1)^(n-1) is the SAME as (-1)^(n+1)....can you prove that really quickly? (Hint: (-1)^2 = 1.)
The second problem is a RECURSIVE formula. Recursive formulas must give you two things:
1) A "base case" (or several base cases) to tell you where you're starting. In this case they give you a(1) and a(2) as your base cases.
2) A recurrence. This gives the NEXT term in terms of some of the preceding terms. In this case, they give you a(n) in terms of a(n-1) and a(n-2).
The reason that they say that a(n) = a(n-1) - a(n-2) ONLY for a>=3 is that if you plug in n=2, you get:
a(2) = a(1) - a(0)
But there IS no 0th term in this sequence! the formula won't work. Thiings get worse if you plug in n=1:
For the first question, you just plug in k=1, and you get (-1)^2 * 1, which is 1.
Then you plug in k=2, and you get (-1)^3 * 2, which is -2.
Keep plugging in all the k's until you get to k=6.
Then what they're wanting you to do is write the sum out, so you write 1 + -2 + ...
You'll have six terms total in your sum.
One hint: whenever you see (-1)^n or (-1)^(n+1), it acts like flipping a lightswitch. (-1)^n will be -1 for odd n, and will be +1 for even n, so your odd terms will be -, and your even terms will be +. Likewise, (-1)^(n+1) will do the opposite: odd terms positive, even terms negative. (-1)^(n-1) is the SAME as (-1)^(n+1)....can you prove that really quickly? (Hint: (-1)^2 = 1.)
The second problem is a RECURSIVE formula. Recursive formulas must give you two things:
1) A "base case" (or several base cases) to tell you where you're starting. In this case they give you a(1) and a(2) as your base cases.
2) A recurrence. This gives the NEXT term in terms of some of the preceding terms. In this case, they give you a(n) in terms of a(n-1) and a(n-2).
The reason that they say that a(n) = a(n-1) - a(n-2) ONLY for a>=3 is that if you plug in n=2, you get:
a(2) = a(1) - a(0)
But there IS no 0th term in this sequence! the formula won't work. Thiings get worse if you plug in n=1:
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