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September 14th, 2009, 04:52 AM   #1
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series

If
and

where n is a positive integer, prove that

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September 14th, 2009, 10:49 AM   #2
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Since S and T are functions of n, I will denote them as S(n) and T(n) respectively.
From the given definitions, S(k) = T(k + 1), where k is any positive integer.
Let U(k) = S(k) + T(k) and V(k) = k(k + 1)(2k + 1)/6, where k is any positive integer,
so that U(1) = 1 + 0 = 1, and V(1) = 1.

U(k + 1) - U(k) = S(k+1) + T(k+1) - (S(k) + T(k))
= S(k + 1) - T(k)
= 1(k+1) + 2(k) + 3(k-1) + ... + r(k+2-r) + (r+1)(k+1-r) ... + (k-1)(3) + k(2) + (k+1)(1)
- 1(k-1) - 2(k-2) - 3(k-3) - ... - r(k-r) - ... - (k-2)(2) - (k-1)(1)
= (k + 1) + (k + 1) + (k + 1) + ... + (k + 1) + (k + 1) (k + 1 times in total)
= (k + 1).

V(k + 1) - V(k) = (k + 1)(k + 2)(2(k + 1) + 1)/6 - k(k + 1)(2k + 1)/6
= (k + 1)(2k + 7k + 6 - (2k + k))/6
= (k + 1)(6k + 6)/6
= (k + 1).

Hence U(k) ? V(k) by induction.
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September 15th, 2009, 03:05 AM   #3
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Re:

Quote:
Originally Posted by skipjack
Since S and T are functions of n, I will denote them as S(n) and T(n) respectively.
From the given definitions, S(k) = T(k + 1), where k is any positive integer.
Let U(k) = S(k) + T(k) and V(k) = k(k + 1)(2k + 1)/6, where k is any positive integer,
so that U(1) = 1 + 0 = 1, and V(1) = 1.

U(k + 1) - U(k) = S(k+1) + T(k+1) - (S(k) + T(k))
= S(k + 1) - T(k)
= 1(k+1) + 2(k) + 3(k-1) + ... + r(k+2-r) + (r+1)(k+1-r) ... + (k-1)(3) + k(2) + (k+1)(1)
- 1(k-1) - 2(k-2) - 3(k-3) - ... - r(k-r) - ... - (k-2)(2) - (k-1)(1)
= (k + 1) + (k + 1) + (k + 1) + ... + (k + 1) + (k + 1) (k + 1 times in total)
= (k + 1).

V(k + 1) - V(k) = (k + 1)(k + 2)(2(k + 1) + 1)/6 - k(k + 1)(2k + 1)/6
= (k + 1)(2k + 7k + 6 - (2k + k))/6
= (k + 1)(6k + 6)/6
= (k + 1).

Hence U(k) ? V(k) by induction.
Thanks a lot for the effort, Skipjack, but I haven't learnt mathematical induction, so is it possible to solve this using knowledge on sequences and series? However, this is my attempt ...

S+T=1(n)+3(n-1)+5(n-2)+7(n-3)+...+2r(2n-2r+1)+...+(2n-1)

how can I continue from here? Thanks !!!
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September 15th, 2009, 04:54 PM   #4
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Instead, why not use the following?

U(n) = U(1) + U(2) - U(1) + U(3) - U(2) + . . . + U(n) - U(n - 1)
= 1 + 2 + . . . + n
= n(n + 1)(2n + 1)/6
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September 16th, 2009, 07:17 AM   #5
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Re:

Quote:
Originally Posted by skipjack
Instead, why not use the following?

U(n) = U(1) + U(2) - U(1) + U(3) - U(2) + . . . + U(n) - U(n - 1)
= 1 + 2 + . . . + n
= n(n + 1)(2n + 1)/6
How did you get this? S+T ??
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September 16th, 2009, 07:47 AM   #6
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Yes, I defined U as S + T.
If you wish, you can also do
V(n) = V(1) + V(2) - V(1) + . . . + V(n) - V(n - 1)
= 1 + 2 + . . . + n.
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