Anyone know a generalization for the sum of the reciprocals of 4 consecutive triangular numbers? (like 1/6 + 1/10 + 1/15 + 1/21 or 1/[(n)(n+1)] + .... 1/[(n+3)(n+4)]
With 3 fractions, I simplified it down to 6/[n(n+3)] but 4 came out to a bunch o big numbers...
On that particular problem, its not in the the form you suggested.
1/6 + 1/10 + 1/15 + 1/21
Its 1/(2)(3) + 1/(2)(5) + 1/(3)(5) + 1/(3)(7). On this one, you add up the first 3 fractions which is 1/6 + 1/10 + 1/15 = 1/3. Add that to 1/21 and you get 8/21. You see alot of that on NS tests - adding parts that are easily added together first instead of doing it all at once.
I dont really see any pattern otherwise. It cycles through the first 4 primes, but I dont know a pattern that can help you with that.
the division by 2 makes them reciprocals of consec triangular numbers...
The first time i did the algebra, I got a huge cubic and didn't think to find its zeroes with my calc.. I did it today and came up with a simple formula though: 8/[n(n+4)]
so 3 reciprocals of consecutive triangular numbers: 6/[n(n+3)]
I don't really have those memorized... for 263547337 I might convert it to 23547337 then 5547337 ---> 147337 ---> 27337 ---> 3337 ---> 337 ---> remainder is 1 (subtracting large multiples of 6 each time.
I thought this might interest you on divisibility by 6. I don't know if there is an easier way yet but there is definitely a pattern.
If:
n mod 2 = 1 and n mod 3 = 1 then n mod 6 = 1
n mod 2 = 0 and n mod 3 = 2 then n mod 6 = 2
n mod 2 = 1 and n mod 3 = 0 then n mod 6 = 3
n mod 2 = 0 and n mod 3 = 1 then n mod 6 = 4
n mod 2 = 1 and n mod 3 = 2 then n mod 6 = 5
n mod 2 = 0 and n mod 3 = 0 then n mod 6 = 0
I am sure all you wanted was more to memorize but this is pretty fast. Do MOD 2 first and if it is 0 the answer is 0, 2, or 4. If it is 1, the answer is 1, 3, or 5. Now just figure out MOD 3.
