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 June 4th, 2014, 02:16 AM #1 Senior Member   Joined: Nov 2013 Posts: 434 Thanks: 8 sum1 find sum 4+11+30+67+......................8003
 June 4th, 2014, 02:56 AM #2 Math Team   Joined: Apr 2010 Posts: 2,780 Thanks: 361 Could be anything. My guess is $\displaystyle 6 \cdot {22 \choose 4} + {20 \choose 2} + 20 \cdot 4 = 44160$
 June 4th, 2014, 02:59 AM #3 Senior Member   Joined: Apr 2014 From: Glasgow Posts: 2,150 Thanks: 730 Math Focus: Physics, mathematical modelling, numerical and computational solutions Do you know how the sum is defined?
 June 4th, 2014, 03:08 AM #4 Global Moderator   Joined: Dec 2006 Posts: 20,480 Thanks: 2039 If the nth term is n³ + 3, the sum is 44160.
 June 4th, 2014, 03:24 AM #5 Math Team   Joined: Apr 2010 Posts: 2,780 Thanks: 361 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 $\displaystyle 6 \cdot {n+3 \choose 4} + {n + 2 \choose 2} + 3 \cdot 20 = 44160$
 June 4th, 2014, 12:10 PM #6 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,329 Thanks: 1024
 June 4th, 2014, 01:21 PM #7 Math Team   Joined: Apr 2010 Posts: 2,780 Thanks: 361 Even that sequence in OEIS ?!

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