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$a_n=a_{n-1}+1/a_{n-1}$ Where, n>1 $a_1=1$ Prove that: $12 < a_{75} < 15$

Note by Kïñshük Sïñgh
4 years, 2 months ago

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Here we go, $a_{n}=a_{n-1}+1/a_{n-1}$

Squaring both sides: $a_{n}^{2}=a_{n-1}^{2}+1/a_{n-1}^{2}+2$

So, if we remove the term: $1/a_{n-1}^{2}$ from RHS

Then, we can say that: $a_{n}^{2}>a_{n-1}^{2}+2$

Therefore, applying a=1,2,3..till $a_{75}$ and $a_{1} =1$ $a_{2}^{2}>3$ $a_{3}^{2}>5$ .... .... ... Series is $2n-1$

Therefore, $a_{75}^{2}>149$ $a_{75}>12$

Now, lets do another part: $a_{n}^{2}=a_{n-1}^{2}+1/a_{n-1}^{2}+2$

As, we know that : Maximum value of $1/a_{n-1}^{2}=1$ Therefore, if we remove this and add 1 then, we can say that: $a_{n}^{2}≤a_{n-1}^{2}+3$

Note that equals to sign will be for n=2 only

Now, $a_{2}^{2}≤4$ $a_{3}^{2}<7$ $a_{4}^{2}<10$ ... .... ..... Series is 3n-2 type:

Therefore, $a_{75}^{2}<223$ $a_{75}<15$

Hence, #$12<a_{75}<15$

- 4 years, 2 months ago

Good solution

- 4 years, 2 months ago

Thanks :)

- 4 years, 2 months ago

Is this question asked by your sir or what.This question was also asked by my friends in school.

- 4 years, 2 months ago

I tried this... But finally i got so many equations... Which were very complex

- 4 years, 2 months ago

No, i go through this question while surfing on Internet

- 4 years, 2 months ago

This question's solution can be found here

- 4 years, 2 months ago

Sorry but.... Solution is not there.. Check it again plz

- 4 years, 2 months ago

Dont worry, I will try to add a solution there

- 4 years, 2 months ago