Question

limit n->infinity
(1²-2²+3²-4²+5²+....n terms) /(n²)
evaluate it plz :'(

Answer 1

didn't get it :'(

Answer 2

Wait, something's not right about that...

Answer 3

answer is 1/2

Answer 4

and -1/2

Answer 5

they have 2 cases in the solution 1)odd 2)even

Answer 6

A little work shows that

\[\sum_{i=1}^{n}\frac{(-1)^{i+1}i^2}{n^2}=(-1)^{n+1}\frac{n+1}{2n}\]

Answer 7

@Zarkon :/

Answer 8

sum up over the odds and evens and them combine

Answer 9

@zarkon didn't get it :'(

Answer 10

@hartnn -can u help me with that ?

Answer 11

are you running the limit to see if the series diverges?

Answer 12

well actually i know only this (1²-2²)+(3²-4²)+...+(n-1)²-n²
why did they wrote (n-1)²-n² that's waht m not getting

Answer 13

do you know induction?

Answer 14

hmm no

Answer 15

wait mathematical induction ?

Answer 16

yeah i know :)

Answer 17

I think im confused about what you are asking.

Answer 18

try doing it without induction.

Answer 19

i don't knwo how to solve it :'(

Answer 20

hint- a^2-b^2=(a+b)(a-b)

Answer 21

okay then what to do
they have taken 2 cases nah

Answer 22

in answer they have given 2 cases 1 for even and 1 for odd
in 1st case :-1/2
and in another
1/2
tehy are gettung

Answer 23

If you take difference of two squares by pairing terms, you will end up with a nice arithmetic series. For n even, every even numbered term in the sequence will have a negative sign in front of it so n^2, the last term will have a negative in front.
\[(1^2 - 2^2) + (3^2 - 4^2) +(5^2 - 6^2) +...+[(n-1)^2 - n^2] =\]
\[(1+2)(1-2) + (3+4)(3-4) +(5+6)(5-6)+...+(n - 1 +n)(n - 1-n) \]
Which incidentally simplifies to \[-3-7-11-...-(2n-1)\]
This is an arithmetic series. For n terms, n even, since we paired two terms to make 1, there are actually n/2 terms. This is what we put into the formula
\[S(n) = \frac{ n }{ 4 }(-3-(2n-1)) \]
Then simplify, divide by n^2 and solve the limit (which is an easy case)

Now try for n odd. The difference will be n^2 will be positive leading. and you will have n^2 outside of the pair.
\[[(1^2 -2^2) + (3^2 - 4^2) +...+[(n-2)^2 - (n-1)^2]] + n^2\]
Find the series sum equivalent for the first n-1 terms (which condense into (n-1)/2 terms like before since we paired them together.and the nth term add it on afterwards.

Answer 24

@mww - thanx ^_^