# Thread: prove each statement by mathematical induction

1. ## prove each statement by mathematical induction

1. $5+10+15+...+5n=\frac{5n(n+1)}{2}$

2. $\frac{4}{5}+\frac{4}{5^2}+\frac{4}{5^3}+...+\frac {4}{5^n}=1-\frac{1}{5^n}$

3. $2^n>n^2$, for $n \ge 5$, $n \varepsilon \mathbb{N}$

4. $5.6+5.6^2+2.6^3+....+5.6^n=6(6^n-1)$

i would appreciate any help so thanks in advance!

2. ## Re: prove each statement by mathematical induction

Do you know how proof by induction works? I.e., the logic behind it and its process?

3. ## Re: prove each statement by mathematical induction

i understand the steps $S_n, S_1, S_k$ and that you assume that $S_k$ is true so that you may prove $S_{k+1}$. I just have trouble proving $S_{k+1}$

4. ## Re: prove each statement by mathematical induction

Hello, tinspire!

$\displaystyle 2.\;\;\frac{4}{5}+\frac{4}{5^2}+\frac{4}{5^3}+ \text{ . . . }+\frac{4}{5^n}\:=\:1-\frac{1}{5^n}$

Verify $S_1\!:\;\;\dfrac{4}{5} \:=\:1- \dfrac{1}{5} \quad\Rightarrow\quad \dfrac{4}{5} \:=\:\dfrac{4}{5}$ . True!

Assume $S_k\!:\;\;\dfrac{4}{5} + \dfrac{4}{5^2} + \dfrac{4}{5^3} + \text{ . . . }+\dfrac{4}{5^k} \;=\;1 - \dfrac{1}{5^k}$

Add $\dfrac{4}{5^{k+1}}$ to both sides:

$\qquad \dfrac{4}{5} + \dfrac{4}{5^2} + \dfrac{4}{5^3} + \text{ . . . }+ \dfrac{4}{5^k} + \dfrac{4}{5^{k+1}} \;=\;1 - \dfrac{1}{5^k} + \dfrac{4}{5^{k+1}}\;\;[1]$

The right side is: $\:1 - \dfrac{1}{5^k} + \dfrac{4}{5^{k+1}} \;=\;1 - \dfrac{5}{5^{k+1}} + \dfrac{4}{5^{k+1}} \;=\; 1 - \dfrac{1}{5^{k+1}}$

Then [1] becomes: $\:\dfrac{4}{5} + \dfrac{4}{5^2} + \dfrac{4}{5^3} +\text{ . . . }+ \dfrac{4}{5^{k+1}} \;=\;1 - \dfrac{1}{5^{k+1}}$

This is $S_{k+1}.$

The inductive proof is complete.

5. ## Re: prove each statement by mathematical induction

Originally Posted by tinspire
1. $5+10+15+...+5n=\frac{5n(n+1)}{2}$

2. $\frac{4}{5}+\frac{4}{5^2}+\frac{4}{5^3}+...+\frac {4}{5^n}=1-\frac{1}{5^n}$

3. $2^n>n^2$, for $n \ge 5$, $n \varepsilon \mathbb{N}$

4. $5.6+5.6^2+2.6^3+....+5.6^n=6(6^n-1)$

i would appreciate any help so thanks in advance!
Originally Posted by tinspire
i understand the steps $S_n, S_1, S_k$ and that you assume that $S_k$ is true so that you may prove $S_{k+1}$. I just have trouble proving $S_{k+1}$
well let's look at the first one

let n=1

$P(1)=[5(1)=\dfrac {5(1)(1+1)}{2}]$

$5\overset{?}{=}5$

Yes. $5=5$ So $P(1)=True$

Now assume $P(n)=True$ and consider $P(n+1)$

$\displaystyle{\sum_{k=1}^{n+1}}5k=$

$5(n+1)+\displaystyle{\sum_{k=1}^{n}}5k=$

$5(n+1)+\dfrac {5n(n+1)}{2}=$

$\dfrac{2\cdot 5(n+1)+5n(n+1)}{2}=\dfrac{5(n+1)(n+2)}{2}=\dfrac{5 (n+1)((n+1)+1)}{2}$

and thus given the statement is true for $n$ it is true for $n+1$

You can work the 3 others. The key is to try and get the $P(n+1)$ expression to include $P(n)$ so you can substitute in the known simplification for $P(n)$.

6. ## Re: prove each statement by mathematical induction

3. 2^n > n^2, for n≥5, nεN

Base case:
n=5, 2^5=32 and 5^2= 25. Hence, lhs > rhs (Base case is true.)

Now, for any k > 5, assume that 2^k > k^2 is true.

Then, we shall prove that 2^(k+1) > (k+1)^2 is also true.

(But first, I am going to digress for a bit.)

2k^2 - (k+1)^2 = 2k^2 - k^2 - 2k -1 = k^2-2k - 1 = (k-1)^2 -2: since k > 5, it's clear that (k-1)^2 - 2 > 0.
My point? for any k > 5, 2k^2 - (k+1)^2 > 0, or better put, 2k^2 > (k+1)^2 starting from 5.

Then, 2^(k+1) = 2*2^k > 2* k^2 = 2k^2 > (k+1)^2 as needed.

7. ## Re: prove each statement by mathematical induction

And for 4, is it supposed to be 5(6+6^2+6^3...+6^n) on the left hand side? [I factored out the 5.]

8. ## Re: prove each statement by mathematical induction

Thanks for everyone's response, I really appreciate all the help and yes I made a mistake. That is supposed to a multipcation sign not a decimal.

9. ## Re: prove each statement by mathematical induction

For the third one don't I have to plug in five for n

10. ## Re: prove each statement by mathematical induction

Oh I didn't realize that's what you did. Never mind

11. ## Re: prove each statement by mathematical induction

In the first step of number 3 for proving S_{k+1} where did 2k^2 come from?

12. ## Re: prove each statement by mathematical induction

What can I do for the fourth one?

13. ## Re: prove each statement by mathematical induction

So that is $5\cdot 6^3$ and not $2\cdot 6^3$ as you have?

Then it is $5(6+ 6^2+ 6^3+ \cdot\cdot\cdot+ 6^n)= 6(6^n- 1)$.

I presume you see that when n= 1 that is 5(6)= 30 and 6(6- 1)= 6(5)= 30

Now if it is true that for n= k, $5(6+ 6^2+ 6^3+ \cdot\cdot\cdot+ 6^k)= 6(6^k- 1)$
then $5(6+ 6^2+ 6^3+ \cdot\cdot\cdot+ 6^k+ 6^{k+1})= 6(6^k- 1)+ 5(6^{k+ 1})= 6^{k+ 1}- 6+ 5(6^{k+1})$

14. ## Re: prove each statement by mathematical induction

yeah thats a five not a two

15. ## Re: prove each statement by mathematical induction

$6\cdot 6^k-6+5(6^k \cdot 6)$

what can i do now?

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