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    The positive integers are bracketed as follows (1),(2,3),(4,5,6),…,where there are r integers in the rth bracketFind expressions for the first and last integers in the rth bracket.
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    (Original post by Carman3)
    The positive integers are bracketed as follows (1),(2,3),(4,5,6),…,where there are r integers in the rth bracketFind expressions for the first and last integers in the rth bracket.
    Well, it's obvious? The 1st bracket has 1 as the first number. The 2nd bracket has 2 as the first number. The 3rd bracket has 4 as the first number. The 4th bracket has 7 as the first number, so the sequence is 1, 2, 4, 7, 11, etc... - if you can spot the closed form for this immediately good job. If not, you can derive it via summing both sides of u_{n+1} - u_n = n subject to u_1 = 1.

    The 1st bracket has 1 as the last number, the second bracket has 3 as the last number. The third bracket has 6 as the last number.

    Can you find a closed form for the sequence that goes: 1, 3, 6, 10, 15, 21, 28? Can you spot it immediately? If so, good. If not: the recurrence relation form is obvious, we have u_{n+1} = u_n + (n+1) starting our indexing from u_1 = 1.

    Then u_2 = 3, u_3 = 3+ (2+1) = 6, u_4 = 6 + (3+1) = 10, \cdots it all seems to work out. If you can't solve that recurrence equation easily, then consider summing both sides of u_{n+1} - u_n = n+1 from k=1 to k=n and finding a nice telescope to get you the explicit formula.

    Of course, this makes it seem like a lot of work, but it's something you should be able to easily do in two minutes or so.
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    (Original post by Zacken)
    Well, it's obvious? The 1st bracket has 1 as the first number. The 2nd bracket has 2 as the first number. [...] The rth bracket has r as the first number.
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    (Original post by Zacken)
    Well, it's obvious? The 1st bracket has 1 as the first number. The 2nd bracket has 2 as the first number. [...] The rth bracket has r as the first number.

    The 1st bracket has 1 as the last number, the second bracket has 3 as the last number. The third bracket has 6 as the last number.

    Can you find a closed form for the sequence that goes: 1, 3, 6, 10, 15, 21, 28? Can you spot it immediately? If so, good. If not: the recurrence relation form is obvious, we have u_{n+1} = u_n + (n+1) starting our indexing from u_1 = 1.

    Then u_2 = 3, u_3 = 3+ (2+1) = 6, u_4 = 6 + (3+1) = 10, \cdots it all seems to work out. If you can't solve that recurrence equation easily, then consider summing both sides of u_{n+1} - u_n = n+1 from k=1 to k=n and finding a nice telescope to get you the explicit formula.

    Of course, this makes it seem like a lot of work, but it's something you should be able to easily do in two minutes or so.
    You could do it like that but there must be an easier way because the question goes on to ask whats the last integer in the 50th bracket and if you used the reccurence relation then that will take long time. Plus i dont think its a reccurence relation question at all as all the other questions in that group arent reccurence related.
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    (Original post by ghostwalker)
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    Darn example sheets... thanks, I'll fix it.
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    (Original post by Carman3)
    You could do it like that but there must be an easier way because the question goes on to ask whats the last integer in the 50th bracket and if you used the reccurence relation then that will take long time. Plus i dont think its a reccurence relation question at all as all the other questions in that group arent reccurence related.
    Like I said, this seems long and hard - but it's really just a matter of working out the pattern. If you can't spot the (fairly obvious) pattern, then you can derive it the way I showed. You're meant to use the recurrence relation to derive a closed form - not plug in 50 numbers into the recurrence...
 
 
 
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