Question

sum

Answer 1

\[\sum_{i=1}^{15}\frac{ 1 }{ n^3 }\left( i - 1 \right)^2\]

Answer 2

(1/n^2) sum (i - 1)^2, i=1..15

Answer 3

\[=\frac{ 1 }{ n^3 }\sum_{i=1}^{15}i^2+\sum_{i=1}^{15}2i+\sum_{i=1}^{15}1\]

Answer 4

1/n^2 times ALL the sum, not just the first sum

Answer 5

I messed up laughing out loud

Answer 6

that is what I meant

Answer 7

why 1/n^2?

Answer 8

maybe set u = i-1,
instead of expanding the square..

Answer 9

man I gotta start learning latex

Answer 10

i meant 1/n^3.

That 3 looks like a 2 in LaTex

Answer 11

how do you do that, @ganeshie8

Answer 12

the the middle sum should be MINUS

Answer 13

-2i yeah

Answer 14

man I hate latex

Answer 15

or instead of expanding like ganeshie8 said, you can just do
Sum i^2, i = 1..14

Answer 16

\(\large \sum \limits_{i=1}^{15}\frac{ 1 }{ n^3 }\left( i - 1 \right)^2 = \sum \limits_{u=1-1}^{15-1}\frac{ 1 }{ n^3 }\left( u \right)^2\)

Answer 17

\[=\frac{ 1 }{ n^3 }\left[ \sum_{i=1}^{15}i^2-\sum_{i=1}^{15}2i+\sum_{i=1}^{15}1\right]\]

Answer 18

okay, how would I evaluate the u^2?

Answer 19

oh I see...

Answer 20

sum of first n squares = n(n+1)(2n+1)/6

Answer 21

thank you, guys
that was a nifty trick I just learned

Answer 22

:) somtimes its hard to see by doing u = i-1 and all.. we cna see it easily by expanding the actual sum a bit :-

\(\large \sum \limits_{i=1}^{15}\frac{ 1 }{ n^3 }\left( i - 1 \right)^2 = \frac{ 1 }{ n^3 } [(1-1)^2 + (2-1)^2 + .... + (15-1)^2]\)

Answer 23

\(\large \sum \limits_{i=1}^{15}\frac{ 1 }{ n^3 }\left( i - 1 \right)^2 = \frac{ 1 }{ n^3 } [(0)^2 + (1)^2 + .... + (14)^2]\)

Answer 24

so, essentially if I had
\[\sum_{i=1}^{5}(2i+1) = \]
ya I tried doing it and I was like okay what's next laughing out loud

Answer 25

now clearly, its a sum of first 14 squares...
we dont need to put u=i-1 and all...( which requires actully thinking in reverse... )

Answer 26

oh.. it has to stay like that i feel... u wana convert it to single sum ha ?
its actually sum of odd numbers from 3 to 11

Answer 27

@sourwing

Answer 28

yeah it starts with 3

Answer 29

so essentially it's 3-11?

Answer 30

3+5+... + 11

just thinking if we can convert it to a single sum..

Answer 31

i mean, in sum notation..

Answer 32

what I have been learning thus far is doing a subtraction so that it turns into 1 - 5 again

Answer 33

what I am actually doing is solving them using theorems

Answer 34

all that k and n notations

Answer 35

yeah we cant simplify the sum notation..
we need to carry out two sums.. .

Answer 36

\(\sum \limits_{i=1}^{5}(2i+1) = \sum \limits_{i=1}^{5}(2i) + \sum \limits_{i=1}^{5}1 \)

Answer 37

u have any better ideas....

Answer 38

that's what I have and I can't use "u"

Answer 39

= 2(n(n+1)/2) + 5

Answer 40

yes got u :)

Answer 41

one last clarification, so going back to the original question
the summation would start from 0 - 14 then?
\[\frac{ 1 }{ n^3 } \sum_{i=0}^{14}i^2\]

Answer 42

I feel like I am not grasping this "u" thing correctly

Answer 43

it's the same thing if you start from 1 to 14

Answer 44

forget u, i also got confused afterwards...
but this looks sum good.... since we verified already that the sum goes : 0^2 + 1^2 + ... + 14^2

Answer 45

*this sum looks good...

Answer 46

then it will be evaluated as
1/n^3 [n(n+1)(2n+1)/6]

Answer 47

1/n^3 [14(14+1)(2*14+1)/6]

Answer 48

the n in 1/n^3 is different from the sum right ?

Answer 49

because it was originally 15?

Answer 50

1015/n^3 is final answer

Answer 51

we dont knw what n in 1/n^3 is, originally..

Answer 52

^^

Answer 53

rem you're right
I assumed that the n is the same n that goes on top of the summation

Answer 54

I thought it should be evaluated along with the sigma properties LAUGHING OUT LOUD omg this is terrible

Answer 55

1/n^3 is like a constant