100
Σ (i^2 - 3j +2)
i=1
how would i figure out the sum?
2007-07-16 07:18:58 · 4 answers · asked by sky l 1 in Science & Mathematics ➔ Mathematics
i meant 3i
sorry about that
2007-07-16 07:30:45 · update #1
for n = 100
∑ i^2 = n(n+1)(2n+1)/6 = 338350
∑ i = n(n+1)/2 = 5050
∑ 1 = n = 100
Answer: 338350 - 3*5050 + 2*100
= 323400
*EDIT*
I'm assuming the "j" was a typo, and you really meant "i"
2007-07-16 07:24:37 · answer #1 · answered by Dr D 7 · 0⤊ 0⤋
What is j? Is it a constant value, or related to i in some way?
Sum (i = 1 to n) i^2 = n(n + 1)(2n + 1) / 6.
Sum (i = 1 to n) i = n(n + 1) / 2.
Sum (i = 1 to n) j = 100*j if j is a constant.
2007-07-16 14:26:37 · answer #2 · answered by Anonymous · 0⤊ 0⤋
100
Σ (i^2 - 3j +2)
i=1
=100
Σ (i^2) +100( - 3j +2)
i=1
You should be able to work it out now.
2007-07-16 14:37:52 · answer #3 · answered by ironduke8159 7 · 0⤊ 0⤋
What is j here?
2007-07-16 14:24:02 · answer #4 · answered by Pareshan Atma 2 · 0⤊ 0⤋