Question

can someone show me hot to evaluate this

Answer 1

its the sum of numbers generated from 0 to n-1

Answer 2

why from zero, its from 1

Answer 3

k=1; but the first term is 1-1 = 0

Answer 4

so what is the process to evaluate this

Answer 5

when k=n; you get the nth number as n-1

Answer 6

what do you mean by "evaluate"? thats a pretty vague term

Answer 7

how do i get the answer

Answer 8

.... depends on what kind of an answer a person is looking for.

Answer 9

i looking for this answer: (n^2-n)/(2)

Answer 10

ok , a formula fir the sum ... thats better

Answer 11

lets write out what we know about the series

A = {0 + 1 + 2 +... + (n-3) + (n-2) + (n-1)}

if we add this to itself we get:

A = {0 + 1 + 2 +...+ (n-3) + (n-2) + (n-1)}
+ A = {{n-1) + (n-2) + (n-3)+...+ 2 + 1 + 0 }
-------------------------------------------------
2A = {(n-1) + (n-1) + (n-1) +...+ (n-1) + (n-1) + (n-1)}

which reduces to:

2A = n(n-1)

A = n(n-1)/2

Answer 12

n(n-1) (n^2-n)
----- = -------
2 2

Answer 13

Does that make sense?

Answer 14

sure does, but i found it a different way

Answer 15

all roads lead to rome, except for the one on the left :)

Answer 16

no, all roads dont lead to rome

Answer 17

whoever told you that is wrong

Answer 18

lets try this. the formula for
\[\sum_{k=1}^n k = \frac{n(n+1)}{2}\] and
\[\sum_{k=1}^n 1=n\] so you can compute
\[\frac{n(n+1)}{2}-n\] for your answer

Answer 19

right thank you, that was the way i did it too

Answer 20

whew!! good thing sate came along and gave us the right way to do it ... ;)

Answer 21

yeah, good thing you suck

Answer 22

now that was uncalled for.

Answer 23

no, your smart *** reply was uncalled for and you know it

Answer 24

you appear to be misinterpreting my response ... but whatever, I gave sate props for his reply.

Answer 25

yeah, yeah sure thats what you did