What kind of sequence is this?
1*2+2*3+3*4+....+2013*2014
1*2+2*3+3*4+....+2013*2014
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Well,
Neither.
the sum S = 1*2+2*3+3*4+....+2013*2014
is a FINITE sum then can be easily computed : this is one one the good points with computers (or Excel macros)
comment from a physical :
2013 is ~ 2*10^3 so 2013*2014 is ~ 2*10^(6000)
and the astrophysician evaluate to 10^80 the number of material atoms in the UNIVERSE...
means S is really... really... a fictional number...
michael
Neither.
the sum S = 1*2+2*3+3*4+....+2013*2014
is a FINITE sum then can be easily computed : this is one one the good points with computers (or Excel macros)
comment from a physical :
2013 is ~ 2*10^3 so 2013*2014 is ~ 2*10^(6000)
and the astrophysician evaluate to 10^80 the number of material atoms in the UNIVERSE...
means S is really... really... a fictional number...
michael
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To find the sum
It's the sum of i*(i + 1)= i^2+i
The sum of i^2 from 1 to n is n(n+1)(2n+1)/6
The sum of i from 1 to n is n(n+1)/2
Adding the two together you can factor out n(n+1) to get
N(n+1)[(2n+1)/6 + 1/2]= n(n+1)[(2n+1)/6+3/6]
n(n+1)(2n+4)/6= n(n+1)(n+2)/3
Now plug in 2013 for n to get the sum
It's the sum of i*(i + 1)= i^2+i
The sum of i^2 from 1 to n is n(n+1)(2n+1)/6
The sum of i from 1 to n is n(n+1)/2
Adding the two together you can factor out n(n+1) to get
N(n+1)[(2n+1)/6 + 1/2]= n(n+1)[(2n+1)/6+3/6]
n(n+1)(2n+4)/6= n(n+1)(n+2)/3
Now plug in 2013 for n to get the sum
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Neither. An arithmetic sequence has a constant difference. A geometric sequence has a constant multiplier. The sequence 2 + 6 + 12 + 20 + 60 + ...does not match either definition