Olympiad Problem 4

Algebra Level 5

The sequence x n {x}_{n} is defined by x 1 = 1 2 , x k + 1 = x k 2 + x k . x_1=\dfrac{1}{2},~~~ x_{k+1} = x_{k}^2 + x_k. If S = 1 x 1 + 1 + 1 x 2 + 1 + 1 x 3 + 1 + + 1 x 100 + 1 S=\dfrac{1}{{x}_{1}+1}+\dfrac{1}{{x}_{2}+1} + \dfrac{1}{{x}_{3}+1} +\cdots + \dfrac{1}{{x}_{100}+1} then find the value of S \left\lfloor S\right\rfloor .


The answer is 1.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Ronak Agarwal
Aug 21, 2014

We have : S = r = 1 1 x r + 1 S=\sum _{ r=1 }^{ \infty }{ \frac { 1 }{ { x }_{ r }+1 } }

We will add and subtract 1 x 1 \frac { 1 }{ { x }_{ 1 } } to S. So we will write :

S= ( 1 x 1 ) + ( 1 x 1 + 1 x 1 + 1 + . . . . . . . . ) (\frac { 1 }{ { x }_{ 1 } } )+(-\frac { 1 }{ { x }_{ 1 } } +\frac { 1 }{ { x }_{ 1 }+1 } +........) Step 1

= 1 x 1 + ( 1 x 1 ( 1 + x 1 ) + 1 ( x 2 + 1 ) + . . . . . . . ) =\frac { 1 }{ { x }_{ 1 } } +(\frac { -1 }{ { x }_{ 1 }(1+{ x }_{ 1 }) } +\frac { 1 }{ { (x }_{ 2 }+1) } +.......) Step 2

Using x k + 1 = x k 2 + x k {x}_{k+1}={x}_{k}^{2}+{x}_{k} we get :

S = 1 x 1 + ( 1 x 2 + 1 ( x 2 + 1 ) + . . . . . . . ) S=\frac { 1 }{ { x }_{ 1 } } +(\frac { -1 }{ { x }_{ 2 } } +\frac { 1 }{ { (x }_{ 2 }+1) } +.......) Step 3

Repeating the same pattern we get

S = 1 x 1 + ( 1 x 3 + 1 ( x 3 + 1 ) + . . . . . . . ) S=\frac { 1 }{ { x }_{ 1 } } +(\frac { -1 }{ { x }_{ 3 } } +\frac { 1 }{ { (x }_{ 3 }+1) } +.......) Step 4

Repeating it till infinity we get :

S = 1 x 1 lim n 1 x n = 1 x 1 S=\frac { 1 }{ { x }_{ 1 } } -\lim _{ n\rightarrow \infty }{ \frac { 1 }{ { x }_{ n } } } =\frac { 1 }{ { x }_{ 1 } } Step 5

S = 1 x 1 \Rightarrow S=\frac { 1 }{ { x }_{ 1 } } Step 6

Put the value of x 1 {x}_{1} to get :

S = 2 S=2 Hence it is clear that the sum of finite terms of this series is less than 2. Hence the integral part of the series is 1.

Why do you add and subtract 1 / x 1/x ?

Trevor Arashiro - 6 years, 9 months ago

Log in to reply

To take advantage of repeating pattern and simplifying the series.

Ronak Agarwal - 6 years, 9 months ago

Log in to reply

So how do you get rid of the +1 in the denominator?

Trevor Arashiro - 6 years, 9 months ago

Log in to reply

@Trevor Arashiro In which step.

Ronak Agarwal - 6 years, 9 months ago

Log in to reply

@Ronak Agarwal Steps 2-3. It looks like you multiple the denominator, by (x+1), but wouldn't you have to multiply the numerator by (x+1)as well.

Trevor Arashiro - 6 years, 9 months ago

Log in to reply

@Trevor Arashiro Actually I have jumped some simple steps here:

What I have done is :

1 x 1 + 1 x 1 + 1 = ( x 1 + 1 ) x 1 ( x 1 + 1 ) + x 1 x 1 ( x 1 + 1 ) = x 1 1 + x 1 x 1 ( 1 + x 1 ) = 1 x 1 ( 1 + x 1 ) \frac { -1 }{ { x }_{ 1 } } +\frac { 1 }{ { x }_{ 1 }+1 } =\frac { -({ x }_{ 1 }+1) }{ { x }_{ 1 }({ x }_{ 1 }+1) } +\frac { { x }_{ 1 } }{ { x }_{ 1 }({ x }_{ 1 }+1) } =\frac { -{ x }_{ 1 }-1+{ x }_{ 1 } }{ { x }_{ 1 }(1+{ x }_{ 1 }) } =\frac { -1 }{ { x }_{ 1 }(1+{ x }_{ 1 }) }

and that's how I got from step 1 to step 2.

Hope this clears things a bit.

Ronak Agarwal - 6 years, 9 months ago

Log in to reply

@Ronak Agarwal I want to clarify one more step here. Here also I have jumped some simple steps. What I have done is :

1 x 1 ( 1 + x 1 ) = 1 x 1 + x 1 2 \frac { -1 }{ { x }_{ 1 }(1+{ x }_{ 1 }) } =\frac { -1 }{ { x }_{ 1 }+{ x }_{ 1 }^{ 2 } }

Now since it is given that x k + 1 = x k 2 + x k {x}_{k+1}={ x }_{ k }^{ 2 }+{ x }_{ k } , Put k = 1 k=1 to get :

x 2 = x 1 2 + x 1 { x }_{ 2 }={ x }_{ 1 }^{ 2 }+{ x }_{ 1 } using this we get:

1 x 1 + x 1 2 = 1 x 2 \frac { -1 }{ { x }_{ 1 }+{ x }_{ 1 }^{ 2 } } =\frac { -1 }{ { x }_{ 2 } }

And this is how we got from step2 to step3.

Ronak Agarwal - 6 years, 9 months ago

@Ronak Agarwal Ahhhh, now I see. Thank you so much for this explanation.

Trevor Arashiro - 6 years, 9 months ago
Jatin Yadav
Sep 13, 2014

x k + 1 = x k ( x k + 1 ) x_{k+1} = x_{k}(x_{k}+1)

1 x k + 1 = x k x k + 1 = x k 2 x k x k + 1 \Rightarrow \dfrac{1}{x_{k}+1} = \dfrac{x_{k}}{x_{k+1}} = \dfrac{x_{k}^2}{x_{k}x_{k+1}}

= x k + 1 x k x k x k + 1 = 1 x k 1 x k + 1 \dfrac{x_{k+1} - x_{k}}{x_{k}x_{k+1} }= \dfrac{1}{x_{k}} - \dfrac{1}{x_{k+1}}

k = 1 n 1 x k + 1 \displaystyle \sum_{k=1}^{n} \dfrac{1}{x_{k}+1}

= k = 1 n ( 1 x k 1 x k + 1 ) \displaystyle \sum_{k=1}^{n} \bigg( \dfrac{1}{x_{k}} - \dfrac{1}{x_{k+1}} \bigg)

= 1 x 1 1 x n \dfrac{1}{x_{1}} - \dfrac{1}{x_{n}}

Clearly, as the sequence goes to \infty at k = k = \infty ,

S = k = 1 n 1 x k + 1 k = 1 1 x k + 1 = 2 0 = 2 \displaystyle \sum_{k=1}^{n} \dfrac{1}{x_{k}+1} \leq \displaystyle \sum_{k=1}^{\infty} \dfrac{1}{x_{k}+1} = 2 - 0 = 2

Hence the answer is 2.

Note : This is an increasing sequence, and x 3 = 9 / 16 + 3 / 4 > 1 x_{3} = 9/16 + 3/4 > 1 , hence, x 100 > 1 , 1 x 100 < 1 x_{100} > 1, \Rightarrow \frac{1}{x_{100}} < 1 ,and the sum is greater than 1.

I think in the second line it should be 1 x k + 1 \frac{1}{x_{k}+1} instead of x k + 1 x_{k}+1 .

Ishan Singh - 6 years, 9 months ago

Log in to reply

Yes, edited. Thanks :)

jatin yadav - 6 years, 9 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...