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**Mario23****Member**- Registered: 2017-01-03
- Posts: 27

Hello ! I've been revising for an exam lately and I came over this problem :

Find the first 3 decimals of sqrt(49n^2+0,35n) where n is a natural number different from 0.I would like a hint on how to do this because it is the first time I see something like it.

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**Mathegocart****Member**- Registered: 2012-04-29
- Posts: 1,884

Can you specify which value of n you mean? The integers are not specific enough, perhaps you meant a sum or limit?

And I assume it is common parlance in Europe that ,'s indicate decimals?

The integral of hope is reality.

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**Mario23****Member**- Registered: 2017-01-03
- Posts: 27

There is no specific value,n is just a positive integer(n>0).I will rewrite the exercise with math symbols :

And yes,that notation is used here for decimals.

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,236

hi

I did a quick 'experiment' with n = 1,2,3 ...10 and found the result always had a whole numbered part that is easy to compute from n, and the same decimal digits for the first three places. [bobbym: I did some EM. Hope you arte proud of me.]

So an answer is easy to calculate. But how to prove it? Working on it .......

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,236

Here we go......

Then you can use the general binomial theorem https://en.wikipedia.org/wiki/Binomial_ … al_theorem

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**Mario23****Member**- Registered: 2017-01-03
- Posts: 27

Ok so from that point I applied that theorem but what should I do next?

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,236

Use the theorem to write down the first three terms of the expansion. Show that term two is independent of n and that later terms are sufficiently small that they will only affect the fourth decimal place and beyond. So you will have an approximation for the expression that is valid to 3dp.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**Mario23****Member**- Registered: 2017-01-03
- Posts: 27

So I should prove that

is independent of n?

EDIT:That's the only thing I can't prove so I could use a hand

*Last edited by Mario23 (2017-01-04 03:08:03)*

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**thickhead****Member**- Registered: 2016-04-16
- Posts: 1,086

Mario23 wrote:

So I should prove that

is independent of n?

EDIT:That's the only thing I can't prove so I could use a hand

It should be**7n*{1/(280n)-1/(8*140^2*n^2)}=1/(40)-7/(8*140^2*n)**

**{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha{Gods rejoice at those places where ladies are respected.}**

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**Mario23****Member**- Registered: 2017-01-03
- Posts: 27

Yeah,I managed to finish it from that point.Thank you all for your help!

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