# Math Help - limit series question..

1. ## limit series question..

$
\lim _{x->\infty}\frac{1^2+3^2+..+(2n-1)^2}{2^2+4^2+..+(2n)^2}
$

i wanted to make a total sum of each series but there is no constant gap
between two members 1 4 9 ..
i cant make an expression of a sum
for the numerator and denominator

??

2. Originally Posted by transgalactic
$
\lim _{x->\infty}\frac{1^2+3^2+..+(2n-1)^2}{2^2+4^2+..+(2n)^2}
$

i wanted to make a total sum of each series but there is no constant gap
between two members 1 4 9 ..
i cant make an expression of a sum
for the numerator and denominator

??
Using standard formulae and a little bit of basic algebra:

$\sum_{k=1}^{n} (2k - 1)^2 = \sum_{k=1}^{n} (4k^2 - 4k + 1) = 4 \left( \sum_{k=1}^{n} k^2 \right) - 4 \left( \sum_{k=1}^{n} k \right) + n = \frac{n(2n-1)(2n+1)}{3}$.

$\sum_{k=1}^{n} (2k)^2 = 4 \sum_{k=1}^{n} k^2 = \frac{2n(n+1)(2n+1)}{3}$.

3. lt (n tending to infinity)[{1^2+2^2+3^2+........+(2n)^2}/{2^2+4^2+......+(2n)^2}-1]

now,1^2+2^2+3^2+......+(2n)^2=2n*(2n+1)*(4n+1)/6.

&,2^2+4^2+......+(2n)^2=2^2{1^2+2^2+3^2+.....+n ^2}
=2^2*n*(n+1)*(2n+1)/6

hence,lt(n--->infinity)[(2*2*4)/(4*1*2)-1]

=1.

4. Originally Posted by mr fantastic
Using standard formulae and a little bit of basic algebra:

$\sum_{k=1}^{n} (2k - 1)^2 = \sum_{k=1}^{n} (4k^2 - 4k + 1) = 4 \left( \sum_{k=1}^{n} k^2 \right) - 4 \left( \sum_{k=1}^{n} k \right) + n = \frac{n(2n-1)(2n+1)}{3}$.

$\sum_{k=1}^{n} (2k)^2 = 4 \sum_{k=1}^{n} k^2 = \frac{2n(n+1)(2n+1)}{3}$.
i didnt study this form can you explain this in simpler terms??

without the $\Sigma$
i understand this but i need to solve such thing in a test
and will not be accepted

5. Originally Posted by transgalactic

i understand this but i need to solve such thing in a test
and will not be accepted
I'm sorry but this is false.

Notation $\sum$ is very useful and I think you should take advantage of.

Having $\sum_{k=1}^3k=1+2+3,$ and that's all. - Basic formulae could be found in Google.

6. Originally Posted by transgalactic
i didnt study this form can you explain this in simpler terms??

without the $\Sigma$
i understand this but i need to solve such thing in a test
and will not be accepted
Originally Posted by transgalactic
$
\lim _{x->\infty}\frac{1^2+3^2+..+(2n-1)^2}{2^2+4^2+..+(2n)^2}
$

i wanted to make a total sum of each series but there is no constant gap
between two members 1 4 9 ..
i cant make an expression of a sum
for the numerator and denominator

??
I, clearly wrongly, assumed from this that you were familiar with the results for $\sum_{k=1}^n k$ and $\sum_{k=1}^n k^2$ and it was simply the 'constant gap' that was causing the problem.

If that's not the case, then I don't see how you could do the question, in which case I don't see why you would be given it in the first place.

7. i can understand how you get this transition

$
4 \left( \sum_{k=1}^{n} k^2 \right) - 4 \left( \sum_{k=1}^{n} k \right) + n = \frac{n(2n-1)(2n+1)}{3}
$

8. Originally Posted by transgalactic
i can understand how you get this transition

$
4 \left( \sum_{k=1}^{n} k^2 \right) - 4 \left( \sum_{k=1}^{n} k \right) + n = \frac{n(2n-1)(2n+1)}{3}
$

9. there are sums of n members +n members
how you get 4 members on the right side.

10. Originally Posted by transgalactic
there are sums of n members +n members
how you get 4 members on the right side.

$\sum_{k=1}^n k = \frac{n(n+1)}{2}$

$\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$

It is basic algebra (or should be) to substitute these expressions into $4 \left( \sum_{k=1}^{n} k^2 \right) - 4 \left( \sum_{k=1}^{n} k \right) + n$ and get $\frac{n(2n-1)(2n+1)}{3}$.

11. i cant see a pattern in your formulas which were found in google
??

i know that the sum of number
from 1 to n

is (n+1)*n/2 so the sum of k is correct

but the k^2 case doesnt hold for it
??

12. Originally Posted by transgalactic
i cant see a pattern in your formulas which were found in google
??

i know that the sum of number
from 1 to n

is (n+1)*n/2 so the sum of k is correct

but the k^2 case doesnt hold for it
??
The formula I have given for the sum of the square of the first n integers is correct. Are you now trying to derive that formula? I really don't understand what your problem is now .... As far as I can see the question you originally posted has been answered ....