Waste less time on Facebook — follow Brilliant.
×

Mind twisting... Try it

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

Note by Kïñshük Sïñgh
2 years, 8 months ago

No vote yet
1 vote

Comments

Sort by:

Top Newest

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, #\[12<a_{75}<15\] Kïñshük Sïñgh · 2 years, 8 months ago

Log in to reply

@Kïñshük Sïñgh Good solution Ronak Agarwal · 2 years, 8 months ago

Log in to reply

@Ronak Agarwal Thanks :) Kïñshük Sïñgh · 2 years, 8 months ago

Log in to reply

Is this question asked by your sir or what.This question was also asked by my friends in school. Ronak Agarwal · 2 years, 8 months ago

Log in to reply

@Ronak Agarwal I tried this... But finally i got so many equations... Which were very complex Kïñshük Sïñgh · 2 years, 8 months ago

Log in to reply

@Ronak Agarwal No, i go through this question while surfing on Internet Kïñshük Sïñgh · 2 years, 8 months ago

Log in to reply

This question's solution can be found here Dinesh Chavan · 2 years, 8 months ago

Log in to reply

@Dinesh Chavan Sorry but.... Solution is not there.. Check it again plz Kïñshük Sïñgh · 2 years, 8 months ago

Log in to reply

@Kïñshük Sïñgh Dont worry, I will try to add a solution there Dinesh Chavan · 2 years, 8 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...