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June 4th, 2014, 02:16 AM   #1
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sum1

find sum
4+11+30+67+......................8003
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June 4th, 2014, 02:56 AM   #2
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Could be anything.
My guess is $\displaystyle 6 \cdot {22 \choose 4} + {20 \choose 2} + 20 \cdot 4 = 44160$
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June 4th, 2014, 02:59 AM   #3
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Do you know how the sum is defined?
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June 4th, 2014, 03:08 AM   #4
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If the nth term is n³ + 3, the sum is 44160.
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June 4th, 2014, 03:24 AM   #5
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The first differences are 7, 19, 37, which are elements of this sequence:
A003215 - OEIS, which is 1, 7, 19, 37, 61. I much like the sequence.
It can be written as $\displaystyle 6 \cdot {n + 1\choose 2} + 1$ for non-negative integers. From there, the sequence can be written as
$\displaystyle 6 \cdot {n+2 \choose 3} + {n + 1 \choose 1} + 3 = 6 \cdot {n+2 \choose 3} + {n + 1 \choose 1} + 3$. Then, for n = 19, we have 8003, for which the sum is $\displaystyle 6 \cdot {n+3 \choose 4} + {n + 2 \choose 2} + 3 \cdot 20 = 44160$
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June 4th, 2014, 12:10 PM   #6
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The On-Line Encyclopedia of Integer Sequences® (OEIS®)

Enter A084378
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June 4th, 2014, 01:21 PM   #7
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Even that sequence in OEIS ?!
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