Question

Let xn be the n-th term of the sequence:
1/1 , 1/4 , 3/4 , 1/9 , 3/9 , 5/9 , 1/16 , 3/16 , 5/16 , 7/16 , 1/25...

Find the maximal n satisfying xn ≥ 1/10

Answer 1

I'll answer yours if you answer mine

Answer 2

thanks

Answer 3

wait, forget it, you're just gonna say because the brain said so

Answer 4

do you mean\[x_n \leftarrow or\rightarrow xn \]

Answer 5

@iTz_Sid I had a little fight with LaTeX and couldn't do anything... It's the first one.

Answer 6

weird sequence
the numerator are increasing prime and denominator are squares

Answer 7

@sooobored yes...
The pattern is (on the denominator) 1²,2²,3²,and so on... On the numerator is 1,3,5,7... , and the number of terms that appears is the same of the denominator without being squared.

The sum of the numbers with the same denominator will always result in 1.

Answer 8

the problem is that there is no general formula for Xn as a function of n , not that I can think of

Answer 9

http://openstudy.com/study#/updates/58288f0ee4b038c47dd29987

Answer 10

I observed it and realized that
\(1,\\ (1/4)+(3/4) =1,\\ (1/9)+(3/9)+(5/9)=1,\\ (1/16)+(3/16)+(5/16)+(7/16)=1\cdots\)

Answer 11

So, the second group has the form of \(x_1 + x_2=1\)

the third group has the form \(x_1+x_2+x_3=1\)

The forth group is \(x_1+x_2+x_3+x_4=1\)

We can assume that \(1=\sum_{i=1}^n x_i\)

Answer 12

But I don't get the question!! is it \(x_n\geq\dfrac{1}{10}\) or \(x*n \geq \dfrac{1}{10}\)

Answer 13

The denominators in the group change as a sequence of
1, 4, 9, 16, 25 ,...

We can find out the rule of it. It is \(d_n= n^2+2n+1\)
(Hopefully you know how to do it)

Answer 14

For the numerator, we can see that on the sequence 1,3,5,7,9,11,...

\(S_1= 1\\S_2= 1+3\\S_3=1+3+5\\S_4=1+3+5+7\)

so, for "group" of the sequence, the numerator has the form \(N_n= S_n\)

Answer 15

@loser66 the question is a "xn" term... I couldn't do it on latex ;-;

Answer 16

And the sum is = the denominator

Answer 17

We have the formula for it. \(\sum_{i=1}^n odd=n^2\)

So far, we have the formula for any group of the sequence is \(\dfrac{n^2}{n^2+2n+1}\)

Answer 18

However, I am too lazy to derive more. hahaha... You can say that I am suddenly stupid.