Mind twisting... Try it

an=an1+1/an1a_n=a_{n-1}+1/a_{n-1} Where, n>1 a1=1 a_1=1 Prove that: 12<a75<1512 < a_{75} < 15

#NumberTheory

Note by Kïñshük Sïñgh
6 years, 11 months ago

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Comments

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

Ronak Agarwal - 6 years, 11 months ago

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No, i go through this question while surfing on Internet

Kïñshük Sïñgh - 6 years, 11 months ago

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

Kïñshük Sïñgh - 6 years, 11 months ago

Here we go, an=an1+1/an1a_{n}=a_{n-1}+1/a_{n-1}

Squaring both sides: an2=an12+1/an12+2a_{n}^{2}=a_{n-1}^{2}+1/a_{n-1}^{2}+2

So, if we remove the term: 1/an121/a_{n-1}^{2} from RHS

Then, we can say that: an2>an12+2a_{n}^{2}>a_{n-1}^{2}+2

Therefore, applying a=1,2,3..till a75a_{75} and a1=1a_{1} =1 a22>3a_{2}^{2}>3 a32>5a_{3}^{2}>5 .... .... ... Series is 2n12n-1

Therefore, a752>149a_{75}^{2}>149 a75>12a_{75}>12

Now, lets do another part: an2=an12+1/an12+2a_{n}^{2}=a_{n-1}^{2}+1/a_{n-1}^{2}+2

As, we know that : Maximum value of 1/an12=11/a_{n-1}^{2}=1 Therefore, if we remove this and add 1 then, we can say that: an2an12+3a_{n}^{2}≤a_{n-1}^{2}+3

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

Now, a224a_{2}^{2}≤4 a32<7a_{3}^{2}<7 a42<10a_{4}^{2}<10 ... .... ..... Series is 3n-2 type:

Therefore, a752<223a_{75}^{2}<223 a75<15a_{75}<15

Hence, #12<a75<1512<a_{75}<15

Kïñshük Sïñgh - 6 years, 11 months ago

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Good solution

Ronak Agarwal - 6 years, 11 months ago

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Thanks :)

Kïñshük Sïñgh - 6 years, 11 months ago

This question's solution can be found here

Dinesh Chavan - 6 years, 11 months ago

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

Kïñshük Sïñgh - 6 years, 11 months ago

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Dont worry, I will try to add a solution there

Dinesh Chavan - 6 years, 11 months ago
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