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    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!
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    Re: prove each statement by mathematical induction

    Do you know how proof by induction works? I.e., the logic behind it and its process?
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    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 $S_{k+1}$
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    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.
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    Re: prove each statement by mathematical induction

    Quote Originally Posted by tinspire View Post
    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!
    Quote Originally Posted by tinspire View Post
    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 $P(n)$.
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    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.
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    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.]
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    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.
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    Re: prove each statement by mathematical induction

    For the third one don't I have to plug in five for n
    Last edited by tinspire; July 22nd 2014 at 05:58 PM.
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    Re: prove each statement by mathematical induction

    Oh I didn't realize that's what you did. Never mind
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    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?
    Last edited by tinspire; July 22nd 2014 at 06:49 PM.
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    Re: prove each statement by mathematical induction

    What can I do for the fourth one?
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    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})
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    Re: prove each statement by mathematical induction

    yeah thats a five not a two
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    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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