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: 2,161 Thanks: 734 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: 20,968 Thanks: 2217 
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 nonnegative 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: 14,597 Thanks: 1038  
June 4th, 2014, 01:21 PM  #7 
Math Team Joined: Apr 2010 Posts: 2,780 Thanks: 361  Even that sequence in OEIS ?!
