#series fm
42 messages · Page 1 of 1 (latest)
its not 5rt3 because when you find the the sum from 5 to 1
you'd need to sub in 5/3, not 5 bc if you look at the summation, you have 3k on the top
and since k is a natural number (counting number) you cant do that
so what you want to do is find the sum from 6 to 1
then subtract the 6th term
because of that 3k on the top you can only find the sums from term 1 to 3, term 1 to 6, term 1 to 9 etc
so the sum from 6 to 1 would be -2rt3
then for the 6th term you just sub in r = 6 into the rtan60r thing and yeah
I’m so confused 😭
idk how you got that 😔 sorry I’m acc so dumb
so the sum from 6 to 1 would be -6rt3
and the 6th term is just 0
yh its tough to grasp icl but when u do its fine
so you see the rtan60r sum
it starts from 1 and ends at 3k
you know that sum is equal to -krt3
so when k = 1, the sum is -rt3, when k = 2 the sum is -2rt3
-rt3 is the sum from 1 to 3k where k = 1, so from r = 1 to r = 3
and -2rt3 is the sum from r = 1 to r = 3k where k = 2, in other words the sum from r = 1 to r = 6
yeah
so now if you want to find the sum from r = 3n to r = 6
you need to do sum of r = 3n to r = 1 minus r = 6 to r = 1
this first part you just use the result in part b, so -nrt3
now for the second part
it's all the stuff i was on about from here onwards
and you get -2rt3
does that make sense or nah?
wait so why is the sum of r=3n to r=1 -nrt3
the first few terms are
r = 1: rt3
r = 2: -2rt3
r = 3: 0
r = 4: 4rt3
r = 5: -5rt3
r = 6: 0
notice how for the r = 3n terms
so r = 3, r = 6 etc u get 0
then for the first two terms you get -rt3 when you add them
you also get this when adding r = 4 and r = 5
so if two of these pairs i guess give you -2rt3
'n' of these pairs give you -nrt3