How Do I Simplify the Right Side of the Mathematical Induction Equation?

[Hazel]

Original Equation: 5+6+7+...+(n+4)=1/2n(n+9)

Ok, I've tried everything to understand this. I'm just not getting it. I understand everything (n=1, k+1, etc) up until this point: "To continue with proof what must be done?". I know you must simplify the right side, but I don't understand how they simplified it to this:
I'm going to start at the adding k+1 to both sides of equation part.
5+6+7+...+(k+4)+[(k+1)+4]=1/2k(k+9)[(k+1)+4]

Now it says to simplify the right side:
1/2k(k+9)+[(k+1)+4]

It says after simplifying it would be:
1/2(k^2+11k+10).

HOW? I don't get it. Please help me. I bet it's going to be so simple, but I can't find out how.

 

[MarkFL]

I'm going to start at the beginning...we are given to prove by induction the following:

\(\displaystyle \sum_{k=1}^n(k+4)=\frac{n(n+9)}{2}\)

First, we check the base case $P_1$:

\(\displaystyle 1+4=\frac{1(1+9)}{2}\)

\(\displaystyle 5=\frac{10}{2}=5\)

True, so we state the induction hypothesis $P_n$:

\(\displaystyle \sum_{k=1}^n(k+4)=\frac{n(n+9)}{2}\)

As our induction step, we may add $((n+1)+4)=n+5$ to both sides:

\(\displaystyle \sum_{k=1}^n(k+4)+((n+1)+4)=\frac{n(n+9)}{2}+n+5\)

On the left, incorporate the new term into the sum while on the right get a common denominator:

\(\displaystyle \sum_{k=1}^{n+1}(k+4)=\frac{n(n+9)+2(n+5)}{2}\)

On the right, distribute:

\(\displaystyle \sum_{k=1}^{n+1}(k+4)=\frac{n^2+9n+2n+10}{2}\)

Combine like terms:

\(\displaystyle \sum_{k=1}^{n+1}(k+4)=\frac{n^2+11n+10}{2}\)

Factor:

\(\displaystyle \sum_{k=1}^{n+1}(k+4)=\frac{(n+1)((n+1)+9)}{2}\)

We have now derived $P_{n+1}$ from $P_n$ thereby completing the proof by induction.

 

[Hazel]

Ok I see you added n+5 to both sides. I'm going to try this on my exercise to see if I can get it now. Thanks!

- - - Updated - - -

I still do not see have they got 1/2k(k+9)+[k+1)+4] simplified to:
1/2(K^2+11k+10)

 

[MarkFL]

Ok I see you added n+5 to both sides. I'm going to try this on my exercise to see if I can get it now. Thanks!

- - - Updated - - -

I still do not see have they got 1/2k(k+9)+[k+1)+4] simplified to:
1/2(K^2+11k+10)

Okay we begin with:

\(\displaystyle \frac{1}{2}k(k+9)+[(k+1)+4]\)

Let's simplify the second term:

\(\displaystyle \frac{1}{2}k(k+9)+(k+5)\)

Now, in order to add these terms, we need a common denominator:

\(\displaystyle \frac{1}{2}k(k+9)+\frac{1}{2}2(k+5)\)

Now both terms have a denominator of 2, so let's put both numerators all over 2:

\(\displaystyle \frac{k(k+9)+2(k+5)}{2}\)

Now, distribute in both terms in the numerator:

\(\displaystyle \frac{k^2+9k+2k+10}{2}\)

Combine like terms:

\(\displaystyle \frac{k^2+11k+10}{2}\)

See it? :D

 

[Hazel]

Oh thank you so much Mark! I have been on this problem for HOURS! The example never told me I needed a common denominator. You couldn't have explained it better! Thanks so much!

 

[MarkFL]

Oh thank you so much Mark! I have been on this problem for HOURS! The example never told me I needed a common denominator. You couldn't have explained it better! Thanks so much!

Glad to have helped! :D

Just remember, whenever you add two terms, you have to have a common denominator..this rule shall follow you the rest of your days in mathematics. :D

 

[Hazel]

Got it! Thanks. :D

 

[Prove It]

Original Equation: 5+6+7+...+(n+4)=1/2n(n+9)

Ok, I've tried everything to understand this. I'm just not getting it. I understand everything (n=1, k+1, etc) up until this point: "To continue with proof what must be done?". I know you must simplify the right side, but I don't understand how they simplified it to this:
I'm going to start at the adding k+1 to both sides of equation part.
5+6+7+...+(k+4)+[(k+1)+4]=1/2k(k+9)[(k+1)+4]

Now it says to simplify the right side:
1/2k(k+9)+[(k+1)+4]

It says after simplifying it would be:
1/2(k^2+11k+10).

HOW? I don't get it. Please help me. I bet it's going to be so simple, but I can't find out how.

There's no need to have to resort to induction. Your LHS is an arithmetic series with $\displaystyle \begin{align*} a = t_1 = 5 \end{align*}$, last term $\displaystyle \begin{align*} l = t_n = n + 4 \end{align*}$ and common difference $\displaystyle \begin{align*} d = 1 \end{align*}$ so the sum is

$\displaystyle \begin{align*} S_n &= \frac{n}{2} \left( a + l \right) \\ &= \frac{n}{2} \left[ 5 + \left( n + 4 \right) \right] \\ &= \frac{n}{2} \left( n + 9 \right) \end{align*}$

 

[Evgeny.Makarov]

First, we check the base case $P_1$:
...
True, so we state the induction hypothesis $P_n$:

I don't think it is good to use the notation $P_1$ before defining $P_n$. After all, $P_1$ is just an instance of $P_n$ where $n$ is instantiated to 1. The first step of every induction proof should be writing $P_n$ in its general form. This allows correctly stating the induction hypothesis and what needs to be proved in the induction step. Moreover, in some proofs by induction figuring out $P_n$ is the most complicated part.

 

[MarkFL]

I don't think it is good to use the notation $P_1$ before defining $P_n$. After all, $P_1$ is just an instance of $P_n$ where $n$ is instantiated to 1. The first step of every induction proof should be writing $P_n$ in its general form. This allows correctly stating the induction hypothesis and what needs to be proved in the induction step. Moreover, in some proofs by induction figuring out $P_n$ is the most complicated part.

In this case, $P_n$ is given up front, it is what is to be proved.