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Prove by mathematical induction that $5 + 5n \leq {n}^2 $ for all integers $n\geq 6$.

Step 1: Base case

Suppose $n = 6$, hence $5 + 5(6) \leq {6}^2 = 35 \leq 36$

We proved that base case is true as 35 is less than or equal to 36.

Step 2: Induction step

We claim that k is true for some integer more than or equal to 6, therefore $5 + 5k \leq {k}^2$ (*)

We now need to prove that k+1 claim is true and that is $5 + 5(k + 1) \leq {(k + 1)}^2$

I am stuck at this step. Somehow I am unable to sub in my claim k which is the (*) equation into my k+1 equation correctly. Is the step up to this point correct?

I have tried expanding out the k+1 equation for the LHS to get $5k + 10$ but it looks absolutely wrong. If I do it to the RHS, I get $7k + 6$ which looks wrong too although I am able to sub in ${k}^2$ to the RHS equation.

Can someone please tell me how to proceed from here on?

Riccardo
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SunnyBoiz
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3 Answers3

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\begin{align} 5+5(k+1) &= 5+5k + 5 \\ &\le k^2+5 \\ &\le k^2 + 2k+1 \\ &=(k+1)^2 \end{align}

The second last step is due to $5 \le 2k+1$ which is equivalent to $2 \le k$, we know this is true since $k \ge 6$.

Siong Thye Goh
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For the induction step, you assume $5+5k\le k^2$.

Now add $2k+1$ to both sides: $$5+5k+(2k+1)\le (k+1)^2.$$ But $k\ge 6$, so certainly $2k+1\ge 5$.

Hence $5+5(k+1)\le (k+1)^2$.

A. Goodier
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  • Could you explain how you derive the $2k + 1 \geq 5$ more? Why is it that this is $\geq$ 5 and not any other value? Sorry if this sounds silly - I am still trying to understand the entire math induction! – SunnyBoiz May 10 '20 at 09:23
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    Well in fact, it is $\ge 13$ since $k\ge 6$, but we only need it to be $\ge 5$ to get the desired conclusion, that $5+5(k+1)\le (k+1)^2$. – A. Goodier May 10 '20 at 09:24
  • Hmm, if let say n is $\geq$ 3. Then we the value should be 2? – SunnyBoiz May 10 '20 at 09:25
  • Not sure what you mean? We are given $n\ge 6$ in the question - the statement is false for $n=3$. – A. Goodier May 10 '20 at 09:27
  • Sorry it was unclear - what I mean is that if I change the question from $\geq$ 6 to 3 instead. – SunnyBoiz May 10 '20 at 09:28
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    But the question wouldn't work for $n=3$, because the statement is false: $5+5\times 3=20$ but $3^2=9$. – A. Goodier May 10 '20 at 09:33
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    @A.Goodier is correct. It doesn't make sense to "change the question from ≥ 6 to 3 instead." As then you would have a false statement. – auspicious99 May 10 '20 at 14:28
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Alternatively, rewrite $5 + 5n \leq n^2$ as $f(n)=n^2- 5n -5 \ge 0$. Then $ f(n+1)-f(n)=2(n-2) \ge 0 $.

lhf
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