Induction (1 Viewer)

[Makematics]

Hi guys, I need help with an induction.

Use mathematical induction to prove that for all integers n≥1,

√1 + √2 + √3 + ... + √n ≤ [(4n+3)√n] / 6

So far I have done the easy bit...

Step 1: Prove true for n=1
LHS=1
RHS= 7/6
LHS≤RHS
Hence it is true for n=1.

Step 2: Assume true for n=k
i.e. Assume √1 + √2 + √3 + ... + √k ≤ [(4k+3)√k] / 6

Step 3: Prove true for n=k+1
i.e. Prove √1 + √2 + √3 + ... + √k + √(k+1) ≤ [(4k+7)√(k+1)] / 6

(BTW Those symbols are square roots)

 

[Makematics]

Bump

 

[Makematics]

Bump

 

[braintic]

Hi guys, I need help with an induction.

Use mathematical induction to prove that for all integers n≥1,

√1 + √2 + √3 + ... + √n ≤ [(4n+3)√n] / 6

So far I have done the easy bit...

Step 1: Prove true for n=1
LHS=1
RHS= 7/6
LHS≤RHS
Hence it is true for n=1.

Step 2: Assume true for n=k
i.e. Assume √1 + √2 + √3 + ... + √k ≤ [(4k+3)√k] / 6

Step 3: Prove true for n=k+1
i.e. Prove √1 + √2 + √3 + ... + √k + √(k+1) ≤ [(4k+7)√(k+1)] / 6

(BTW Those symbols are square roots)

This is Exercise 8.1 Q7a from Terry Lee. You can look at the solutions at the back of his book.

 

[Sy123]

Hi guys, I need help with an induction.

Use mathematical induction to prove that for all integers n≥1,

√1 + √2 + √3 + ... + √n ≤ [(4n+3)√n] / 6

So far I have done the easy bit...

Step 1: Prove true for n=1
LHS=1
RHS= 7/6
LHS≤RHS
Hence it is true for n=1.

Step 2: Assume true for n=k
i.e. Assume √1 + √2 + √3 + ... + √k ≤ [(4k+3)√k] / 6

Step 3: Prove true for n=k+1
i.e. Prove √1 + √2 + √3 + ... + √k + √(k+1) ≤ [(4k+7)√(k+1)] / 6

(BTW Those symbols are square roots)

Okay, now this is quite a good question because the inequality is very.... 'accurate', as in the bounds are quite close. Note that if the question was designed so that Step 3 ended up with (4k+9) as the bracket, teh question is simple as we simply acknowledge that sqrt(k) < sqrt(k+1)

But the solution I have, is very much derived from the answer. What I did was.
If the expression on the RHS of what we need is correct, then the following inequality holds:



This is true IF Step 3 is true. So that means the problem resolves itself into proving the above inequality. The LHS is from adding sqrt(k+1) to Step 2.

That means:

IF Step 3 is true, then:



Then expanding, then simplifying, will yield, the NEED to prove that:



Now, from the AM-GM inequality:



Setting a and b appropriately:



That means that half of 4k+1 is greater than the RHS, that means 1 of 4k+1 is greater than RHS



Technicalities.

That means, inequality: must hold.

Therefore, we can establish what needed to be true:



Hence, Step 3 is true.

Therefore:

*Induction essay*

Proven by mathematical induction

 

[Trebla]

First result we need is that for positive integer k



Proof: First note that for positive integer k



and the result follows.

Now to the inductive step:

 

[Makematics]

Ah thanks heaps guys @Braintic ah right i got it from cambridge, was unaware that it was also in Terry Lee.
@Sy and @Trebla : Cheers Nice solutions.

 

[Makematics]

was really excited to see the solution

 

[Makematics]

hmm i did this exact method, but my dad told me that the above statement was incorrect :/ i agree that 1 is greater than or equal to 0, although i would like a clear explanation of why it is so. could you explain why it is correct?

 

[Trebla]

hmm i did this exact method, but my dad told me that the above statement was incorrect :/ i agree that 1 is greater than or equal to 0, although i would like a clear explanation of why it is so. could you explain why it is correct?

1 is either {greater than zero} OR {equal to zero}.

In this case we know with certainty that the former is true in a strict sense (1 > 0 is a strict inequality), but the statement of the weak inequality is still true.

 

[Makematics]

hmm yeah that makes sense i guess. Also trebla how do you know that you have to start off with 1 is greater than or equal to 0?

 

[Trebla]

hmm yeah that makes sense i guess. Also trebla how do you know that you have to start off with 1 is greater than or equal to 0?

Work backwards from the inequality you need to prove

 

[Makematics]

oh alright cool