Give you guys some exercise,lol

Stanley_Marsh · 2007-01-20 10:54:21

Find the sum of 1x2x3+2x3x4+3x4x5............

Find the formula for the sequence 2A(n-1) + An = 2^n

A(n-1) and An represent the sequence

mathsyperson · 2007-01-21 06:30:43

I'll do the first one of those.

I think it would be easier to include the term 0x1x2 on the front of that, because that way you can put the terms in the form (n-1)n(n+1). 0x1x2=0, so it doesn't affect the sum, but it does mean that you need to shift the n value up by one. For example, if you wanted to find the sum of the first 5 terms in your sequence, then that's actually the first 6 terms in the altered sequence so you'd put n=6.

Anyway, what we have now is a sequence where the terms are in the form n(n-1)(n+1). This multiplies out and cancels a bit to give n³-n.

Now for the nice bit: It just so happens that

This means that

That's the answer according to my altered sequence, and so to shift it back so that it works for the original, we add 1 to each of the n's:

Test:
1x2x3 + 2x3x4 + 3x4x5 = 6+24+60 = 90.
The sum of the first three terms is given by [(3+1)(3+2)/2]² - [(3+1)(3+2)/2] = 100-10=90. Success!

Stanley_Marsh · 2007-01-24 12:37:34

Correct~ anyone attempt the second question?

John E. Franklin · 2007-01-24 15:33:06

Sweet calculating mathsy!!! You know some nice facts to get that one done that I didn't know.
Fact AA: So (1 + 2 + 3 + 4)^2 is 100, just like 1 + 8 + 27 + 64.
Fact BB: And 1 to 10 added up is 55, which is 5 times 11.
I see where you are coming from, but I didn't know Fact AA.

John E. Franklin · 2007-01-24 15:48:08

Am I right?

Ricky · 2007-01-25 04:27:45

Find the formula for the sequence 2A(n-1) + An = 2^n

Almost, John.

But this is of course assuming the starting values are in this form.

Stanley_Marsh · 2007-01-26 10:18:33

I don't know the answer either ,lol
I divide two sides by 2^n, then A(n-1)/2^(n-1) + An/2^n=1
Denote An/2^n as another sequence Bn , then Bn+B(n-1)=1
you can add a minor to it
B2+B1=1
-B3-B2=-1
.......
IF N is even the last one will be Bn+B(n-1)=1 , add altogether Bn+B1=1 , thus Bn=1-B1
An=2^n - 2^(n-1)*A1
If N is odd , the last one will be -Bn-B(n-1)=-1 , add altogether ,-Bn+B1=0 , thus Bn=B1
An=2^(n-1)*A1

That's what I did ~lol

Math Is Fun Forum

#1 2007-01-20 10:54:21

Give you guys some exercise,lol

#2 2007-01-21 06:30:43

Re: Give you guys some exercise,lol

#3 2007-01-24 12:37:34

Re: Give you guys some exercise,lol

#4 2007-01-24 15:33:06

Re: Give you guys some exercise,lol

#5 2007-01-24 15:48:08

Re: Give you guys some exercise,lol

#6 2007-01-25 04:27:45

Re: Give you guys some exercise,lol

#7 2007-01-26 10:18:33

Re: Give you guys some exercise,lol

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