# Sequences: I hate them!!!Watch

This discussion is closed.
#1
I have a long question which seems to lead to the last bit which I can't understand

The sequence is defined by un+1 = 1 - 1/un

I don't kow if the above has anything to do with it as the following is part d) of the question

Given that u1 = 2, evaluate 1000Sigma i=1 (ui)^2

The answr is: 333 * 21/4 + 4 = 7009/4

I have no idea where all these figures came from and some explanation as to what is going one, would be much appreciated.

Exam next Wednesday: I'll never make it!!!

jorge
0
14 years ago
#2
this part d looks more difficult than it actually is. just write down the sequence for a couple of values and you´ll see:

u1 = 2
u2 = 1/2
u3 = -1
u4 = 2 and now you´ll see that the sequence just repeats itself.

now we know that from u1 to u1000 there are 333 times 2, 1/2 and -1 respectively + one 2. now all you need to do is to square 2, 1/2 and 1 to get the sum asked for in the question:

333 * 2^2 + 333 * (1/2)^2 + 333 * (-1)^2 + 1 * 2^2 = 333 * (4 + 1/4 + 1) + 4 = 7009/4 .
0
14 years ago
#3
I hope i'm not misunderstanding your notation, since its poorly written to say the least. Anyway,

1000Sigma i=1 (ui)^2 = (u_1)^2 + (u_2)^2 + ... + (u_1000)^2

It is given that u_1 = 2. Hence, u_2 = 1-1/2=(1/2). u_3=1-1/(1/2)=-1. u_4=1-1/(-1)=2. Hence, the pattern repeats, i.e. u_(1+3n)=2, u_(2+3n)=1/2, u_(3+3n)=-1

Hence, in the summation expression, we have 334 '2's, 333 '(1/2)s' and 333 (-1)s. Hence, our expression is

334(2^2) + 333(0.5^2) + 333(-1^2)= 1336+83.25+333 = 1752.25 = 7009/4
0
#4
(Original post by J.F.N)
I hope i'm not misunderstanding your notation, since its poorly written to say the least. Anyway,

1000Sigma i=1 (ui)^2 = (u_1)^2 + (u_2)^2 + ... + (u_1000)^2

It is given that u_1 = 2. Hence, u_2 = 1-1/2=(1/2). u_3=1-1/(1/2)=-1. u_4=1-1/(-1)=2. Hence, the pattern repeats, i.e. u_(1+3n)=2, u_(2+3n)=1/2, u_(3+3n)=-1

Hence, in the summation expression, we have 334 '2's, 333 '(1/2)s' and 333 (-1)s. Hence, our expression is

334(2^2) + 333(0.5^2) + 333(-1^2)= 1336+83.25+333 = 1752.25 = 7009/4
Thank you, both of you.

I can see what the question is driving at.

However, I supplied the answer that was in the book. But how am I supposed to arrive at it, if I don't know it?

Sorry for carrying on.

Jorge
0
14 years ago
#5
(Original post by Jorge)
Thank you, both of you.

I can see what the question is driving at.

However, I supplied the answer that was in the book. But how am I supposed to arrive at it, if I don't know it?

Sorry for carrying on.

Jorge
No problem. The best thing you can do when seeing a sigma sign is to actually write out the terms of the summation. This is what I did here:

1000Sigma i=1 (ui)^2 = (u_1)^2 + (u_2)^2 + ... + (u_1000)^2

Now, the problem is to calculate (u_1)^2 + (u_2)^2 + ... + (u_1000)^2

We are given u_1, so the obvious question is: whats u_2? We are given a way to calculate u_2, so we calculate u_2. Similarly, we calculate u_3. And u_4. Obviously, we can't keep on calculating up to u_1000! Hence, we try to find a pattern. The pattern in this case was that 2,0.5,-1 repeating itself, such that u_1, u_4, u_7,... = 2, and u_2, u_5, u_8,... = 0.5, and u_3, u_6, u_9,... = -1. Now, we calculated how many '2's, how many 0.5s and how many -1s existed in this sequence:

(u_1)^2 + (u_2)^2 + ... + (u_1000)^2

From then on, it was just work on the calculator. I hope this helps. Feel free to ask (more specific) questions.
0
#6
(Original post by J.F.N)
No problem. The best thing you can do when seeing a sigma sign is to actually write out the terms of the summation. This is what I did here:

1000Sigma i=1 (ui)^2 = (u_1)^2 + (u_2)^2 + ... + (u_1000)^2

Now, the problem is to calculate (u_1)^2 + (u_2)^2 + ... + (u_1000)^2

We are given u_1, so the obvious question is: whats u_2? We are given a way to calculate u_2, so we calculate u_2. Similarly, we calculate u_3. And u_4. Obviously, we can't keep on calculating up to u_1000! Hence, we try to find a pattern. The pattern in this case was that 2,0.5,-1 repeating itself, such that u_1, u_4, u_7,... = 2, and u_2, u_5, u_8,... = 0.5, and u_3, u_6, u_9,... = -1. Now, we calculated how many '2's, how many 0.5s and how many -1s existed in this sequence:

(u_1)^2 + (u_2)^2 + ... + (u_1000)^2

From then on, it was just work on the calculator. I hope this helps. Feel free to ask (more specific) questions.
Got it. Thanks
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