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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
Numbers are the essence of the Universe
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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!
Why did the vector cross the road?
It wanted to be normal.
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Correct~ anyone attempt the second question?
Numbers are the essence of the Universe
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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.
igloo myrtilles fourmis
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igloo myrtilles fourmis
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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.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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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
Numbers are the essence of the Universe
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