a.myhall
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write down the first 4 terms of the sequences and find a formula for Un.

U1= 2
Un = U(n-1) + n.

I know the first terms are 2, 4, 7, 11, but how can I work out the formula for Un. The answer is 1 + n/2(n + 1)... am I meant to realise this just by looking at the sequence? Thank you
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Pangol
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(Original post by a.myhall)
write down the first 4 terms of the sequences and find a formula for Un.

U1= 2
Un = U(n-1) + n.

I know the first terms are 2, 4, 7, 11, but how can I work out the formula for Un. The answer is 1 + n/2(n + 1)... am I meant to realise this just by looking at the sequence? Thank you
I think you are supposed to realise that the rule is "add two, add three, add four", and so on. Actually formulating it symbolically is a bit of a challenge, but this is the main step.
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Phil_C
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if the 2nd differences go up 1,2,3,4 etc then the sequence can be shown to have the triangular numbers at its core that has the general formula (1/2)n(n+1)

Alternatively notice that the 2nd differenc in the squence is constant, in this case a common difference of 1 and work on the principle that the level of difference that is common (lets call it D) gives the highest power of the polynomial equati0n that will match that sequence of digits.

So with a 2nd difference constant the sequence is goverened by an equation that will be n^D at it's core

A method (not the only one) for this sequence, having identified it as quadratic (common 2nd difference) would be say that the sequence has nth term = an^2 + bn + c
now substitute n = 1, 2, 3, and get linear equations in a, b, c = 2, 4 and 7 and then solve these. use forth term = 11 as your check.
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