from what i'm gathering i would just use the rule of sums and coefficients to isolate and then replace i with the sum of squares, but i can't seem to find an approach to getting n to cancel out and give my limit meaning. am i goofing?
From what i'm gathering i would just use the rule of sums and coefficients to isolate and then replace i with the sum...
Brandon Campbell
Henry Morales
you goofed in that you replaced delta x with 2+3/n.
x_i also isnt 2+3/n but 2+3i/n.
Blake Gutierrez
delta x is 3/n.
x_i is 2 + 3 i /n
plug those into the sum, get sum sums of square... you just need the formula for sum i^2 and sum i.
Jace Myers
ah my bad. ok i think this works better.
Tyler Hughes
you need a pair of parentheses.
\sum_{k=1}^n (4-2*(2+3 k / n ))*3/n = -9-9/n.
As n goes to infinity, the sum goes to -9.
Elijah Jackson
like this?
sorry i didn't mean to try and parrot. it's just that i can't see the simplification approach. i still end up with n and i in my expression. ill show what i got.
Adam Hall
ah sorry the summation doesn't apply there since there's no i. damn i keep making mistakes.
Nathan Wilson
you should really get your arithmetic in check before attempting to do analysis.
the last line is a complete clusterfuck
Sebastian James
at the second to last line you should multiply everything out and then to evaluate the summation you just substitute n(n+1)/2 for i. At least I think. It gave me the right answer anyway.
Andrew Murphy
yes, that's it I think. you need to use 1+2+...+n = n*(n+1)/2 and 1+2^2 + ... + n^2 = n*(n+1)*(2n+1)/6 and algebra.