A few questions about series and sequences
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A few questions about series and sequences

[From: ] [author: ] [Date: 11-04-25] [Hit: ]
you know that a(2) = -1 and a(1) = +1.Thats why you need the base cases!You get:a(3) = -1 - +1 = -2You can then find a(4) by pluggin in n=4.Keep going until you find a(7), and then list all seven terms.-Alrite,......
a(1) = a(-1) - a(0).

But it works fine for n=3, n=4, etc. Plug in n=3 to get:
a(3) = a(2) - a(1).
From your base cases, you know that a(2) = -1 and a(1) = +1. That's why you need the base cases! You get:
a(3) = -1 - +1 = -2

You can then find a(4) by pluggin in n=4. Keep going until you find a(7), and then list all seven terms.

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Alrite, not too bad. So for series, there is obviously a sum of many things. In this case, that is just a notation that gives you how the series should look like in a concise form. To expand this, take those numbers below and above the E and all the numbers in between and plug them into the "function" where every you see a k. Then add up all those numbers you get for each k you pluged in.

In this case, you plug in 1, 2 ,3, 4, 5, and 6 into that function and add up all the values you get when you plug in them. So 1 - 2 + 3 - 4 + 5 - 6 =..... well, u can do that :P

The next thing is about recursive sequence. The idea is that the next term in the sequence is related to the 2 terms before it. For this, take the 2 terms you know and plug them into that recursive forumula. Note that the subscript means which number it is in the sequence.

For this, the 3th number, a3, = 1 - -1 = 2. Then the next term is the -1 - 2 =-3. then just keep going and going.

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1. ∑ (-1)^(k + 1) k = 1 - 2 + 3 - 4 + 5 - 6 = -3

2. a_1 = 1, a_2 = -1; a_n = a_(n - 2) - a_(n - 1)

a_3 = a_1 - a_2 = 1 - (-1) = 2
a_4 = a_2 - a_3 = -1 - 2 = -3
a_5 = a_3 - a_4 = 2 - (-3) = 5
a_6 = a_4 - a_5 = -3 - 5 = -8
a_7 = a_5 - a_6 = 5 - (-8) = 13

1, -1, 2, -3, 5, -8, 13 are the first seven terms.
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