You know that Ln = L(n-1) + L(n-2) for all b
You also know that L0 + ... + Ln = L(n+2) - 1 for some n
Can you now prove it for n+1?
i.e. that
L0 + ... + L(n+1) = L(n+3)
(if the ... in the 0 and 1 case are too confusing to you, then see if you can do the inductive step starting from the n=2 base case, it shouldn't really matter what n you use for the inductive step, since the same proof should work for all n larger than the base case
the range from 0 to 0 is (0), so when n=0, the sum is
L0
the range from 0 to 1 is (0,1) so when n=1, the sum is
L0 + L1
the range from 0 to 2 is (0,1,2) so when n=2 the sum is
L0 + L1 + L2
the range from 0 to 3 is (0,1,2,3) so when n=3 the sum is
L0 + L1 + L2 + L3
0
PMG76Author Commented:
The formula is then n/2 (2a + (n-1) d
The distance is not a constant distance. I know how to do summations, but not making my own or this complicated.
the L0 + L1 + ... + Ln notation, in attempting to illustrate the sequence in general, may be confusing in suggesting that L1 is necessarily part of the sum.
sum(i=0...n : Li) may be less misleading in that sense, but is also more abstract, which may make it less intuitive.
You can just take our word that
L0 + L1 + ... + Ln
means
L0
when n=0
...
and that
L0 + L1 + ... + Ln
means
L0 + L1 + L2 + L3
when n=3
and that
L0 + L1 + ... + Ln
means
L0 + L1 + L2 + L3 + L4
when n=4
but doesn't matter as much whether you believe us for n=0 and n=1
as long as you can extend it to larger n, and can do the inductive step
from some n for which
L0 + L1 + ... + Ln = Ln+2 - 1
is true to the next value of n
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So, how far did you get ?
Can you prove the base case (n = 0) ?
Can you prove the step ?