How to solve the problem

[MathsLearner123]

I am self learning and trying to solve the above problem. I have no clew how to solve it. Any hints how to proceed? I am planning to use Taylor series will it help me?

 

[taeto]

It is possible to use Taylor series to get a proof even in the more general case when n is any real number greater than or equal to 2. But it looks like it will be complicated. Do you know the Taylor expansion of the RHS of your inequality, that is, the expansion of 7^x as a function of x?

It looks more like you are supposed to apply induction on n: en.wikipedia.org/wiki/Mathematical_induction

The easiest might be to first show that the result holds when n = 2, and then prove with simple calculus that the difference LHS - RHS is always increasing for larger n (again assuming the variable is real-valued). This would be kind of "induction on real numbers".

Edited June 17, 2018 by taeto
added missing word

[MathsLearner123]

Thank you for the help. Can I prove some thing like this

For n=2

LHS: 2 * (9 + 16)  = 50 RHS: 7^2 = 49

50 > 49. Difference (50 - 49) = 1;

For n=3;

LHS: 4*(27+64) = 364, RHS: 7 ^ 3 = 343

364 > 343. Differnce (364 - 343) = 21.

The difference is increasing for the increasing n. Hence the equation is valid for all  n >=2.  Is it correct way of solving? Please advise.

 

 

 

 

[taeto]

You are getting somewhere:

1 hour ago, MathsLearner123 said:

For n=2

LHS: 2 * (9 + 16)  = 50 RHS: 7^2 = 49

50 > 49. Difference (50 - 49) = 1;

is essentially the basis of the induction proof. You have shown that the inequality is true when n is equal to 2. This is important. However,

1 hour ago, MathsLearner123 said:

For n=3;

LHS: 4*(27+64) = 364, RHS: 7 ^ 3 = 343

364 > 343. Differnce (364 - 343) = 21.

is not part of the proof by induction. It shows that the inequality is true when \( n \) is equal to 3. The increase in the difference between LHS and RHS is irrelevant. How would you argue that the statement is also true for \( n = 4, \) except by calculating both sides of the inequality for that case as well? You do not know whether the difference between LHS and RHS will suddenly drop enough to make the inequality false for that value of \( n, \) or for some larger value.

Did you read the Wikipedia page, or anything else, on how to prove something by induction?

Do you know how to prove that \( 2^n \geq n^2 \) holds for every \( n > 3 \) by induction? You should be familiar with this baby example if you want to prove anything as complicated as the problem that you posted. I am puzzled if you are asked to prove anything complicated like that without having seen the simplest examples. Where did you get your problem from? Is there anything to suggest that you should actually try to solve it using Taylor series? 

Edited June 17, 2018 by taeto

[MathsLearner123]

There is no suggestion that i should try with Taylor series. As suggested by you i will try solving simpler induction problems then will attempt this problem. I am trying to solve some of the entrance examination papers which are generally considered very tough. 

[MathsLearner123]

 

I have attempted the problem. Can you please advise if i am doing correctly. I want to prove equation 1 again by induction is it correct? Please advise.

Induction.pdf

[taeto]

1 hour ago, MathsLearner123 said:

 

I have attempted the problem. Can you please advise if i am doing correctly. I want to prove equation 1 again by induction is it correct? Please advise.

Induction.pdf

It is a good attempt, though only about half right.

You should not write "for n = n+1", since obviously n is always different from n+1, except in very peculiar situations, which do not appear in ordinary arithmetic. Instead you should write something like "Assume that the statement is true for n, and consider the statement for n+1" (which means, the statement about n+1 that you get after replacing n by n+1 in the original statement).

So you will assume that you have some number n for which you already know \( 2^n \geq n^2 \). The only thing left to show now is \( 2^{n+1} \geq (n+1)^2 \) and you will be done.

You should do it by first grabbing the thing \( 2^{n+1} \) that you want to prove something about (namely that it is \( \geq (n+1)^2 \) ). But you should look at the statement that you know that you are allowed to assume, which is   \( 2^n \geq n^2 \). You have to ask yourself, what does the fact \( 2^n \geq n^2 \) tell me about how large the number  \( 2^{n+1} \) will be? And then, if you can figure this out, how does your new guaranteed smallest possible value for \( 2^{n+1} \), which you just got, compare to the number \( (n+1)^2 \)? Is it a larger number? If it is, then you are done. 

The kind of statement here seems tricky because it involves an inequality instead of just a formula with an equality sign in it. Maybe you should practice a bit more with arithmetical theorems that can be written as equations. Something like \( 1 + 2 + \cdots + n = \frac{n(n+1)}{2} \) is another standard exercise.