The sequence x n is defined by x 1 = 2 1 , x k + 1 = x k 2 + x k . If S = x 1 + 1 1 + x 2 + 1 1 + x 3 + 1 1 + ⋯ + x 1 0 0 + 1 1 then find the value of ⌊ S ⌋ .
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Why do you add and subtract 1 / x ?
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To take advantage of repeating pattern and simplifying the series.
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So how do you get rid of the +1 in the denominator?
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@Trevor Arashiro – In which step.
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@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.
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@Trevor Arashiro – Actually I have jumped some simple steps here:
What I have done is :
x 1 − 1 + x 1 + 1 1 = x 1 ( x 1 + 1 ) − ( x 1 + 1 ) + x 1 ( x 1 + 1 ) x 1 = x 1 ( 1 + x 1 ) − x 1 − 1 + x 1 = x 1 ( 1 + x 1 ) − 1
and that's how I got from step 1 to step 2.
Hope this clears things a bit.
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@Ronak Agarwal – I want to clarify one more step here. Here also I have jumped some simple steps. What I have done is :
x 1 ( 1 + x 1 ) − 1 = x 1 + x 1 2 − 1
Now since it is given that x k + 1 = x k 2 + x k , Put k = 1 to get :
x 2 = x 1 2 + x 1 using this we get:
x 1 + x 1 2 − 1 = x 2 − 1
And this is how we got from step2 to step3.
@Ronak Agarwal – Ahhhh, now I see. Thank you so much for this explanation.
x k + 1 = x k ( x k + 1 )
⇒ x k + 1 1 = x k + 1 x k = x k x k + 1 x k 2
= x k x k + 1 x k + 1 − x k = x k 1 − x k + 1 1
k = 1 ∑ n x k + 1 1
= k = 1 ∑ n ( x k 1 − x k + 1 1 )
= x 1 1 − x n 1
Clearly, as the sequence goes to ∞ at k = ∞ ,
S = k = 1 ∑ n x k + 1 1 ≤ k = 1 ∑ ∞ x k + 1 1 = 2 − 0 = 2
Hence the answer is 2.
Note : This is an increasing sequence, and x 3 = 9 / 1 6 + 3 / 4 > 1 , hence, x 1 0 0 > 1 , ⇒ x 1 0 0 1 < 1 ,and the sum is greater than 1.
I think in the second line it should be x k + 1 1 instead of x k + 1 .
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We have : S = ∑ r = 1 ∞ x r + 1 1
We will add and subtract x 1 1 to S. So we will write :
S= ( x 1 1 ) + ( − x 1 1 + x 1 + 1 1 + . . . . . . . . ) Step 1
= x 1 1 + ( x 1 ( 1 + x 1 ) − 1 + ( x 2 + 1 ) 1 + . . . . . . . ) Step 2
Using x k + 1 = x k 2 + x k we get :
S = x 1 1 + ( x 2 − 1 + ( x 2 + 1 ) 1 + . . . . . . . ) Step 3
Repeating the same pattern we get
S = x 1 1 + ( x 3 − 1 + ( x 3 + 1 ) 1 + . . . . . . . ) Step 4
Repeating it till infinity we get :
S = x 1 1 − lim n → ∞ x n 1 = x 1 1 Step 5
⇒ S = x 1 1 Step 6
Put the value of x 1 to get :
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.