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

In summary: So why not write $P_1$ first? I agree that the notation $P_1$ is redundant, so let's just write the base case as 1+4=1/2(1(1+9))=5. Then we can go on to define $P_n$.
  • #1
Hazel
11
0
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.
 
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  • #2
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.
 
  • #3
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)
 
  • #4
Hazel said:
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
 
  • #5
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!
 
  • #6
Hazel said:
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
 
  • #7
Got it! Thanks. :D
 
  • #8
Hazel said:
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

$\displaystyle \begin{align*} S_n &= \frac{n}{2} \left( a + l \right) \\ &= \frac{n}{2} \left[ 5 + \left( n + 4 \right) \right] \\ &= \frac{n}{2} \left( n + 9 \right) \end{align*}$
 
  • #9
MarkFL said:
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.
 
  • #10
Evgeny.Makarov said:
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.
 

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

What is mathematical induction?

Mathematical induction is a mathematical proof technique used to prove statements about natural numbers or other discrete structures. It involves two steps: the base case, where the statement is shown to be true for the first natural number, and the inductive step, where it is shown that if the statement is true for one natural number, it is also true for the next one.

How is mathematical induction different from other proof techniques?

Mathematical induction is different from other proof techniques, such as direct proof or proof by contradiction, because it relies on the principle of strong induction, which states that if a statement is true for all natural numbers up to some number n, then it is also true for the next natural number, n+1. This allows for the proof of a statement for an infinite number of cases.

What types of statements can be proven using mathematical induction?

Mathematical induction is typically used to prove statements about natural numbers, such as properties of sequences or sums. It can also be used to prove statements about other discrete structures, such as graphs or trees, as long as they have a well-defined ordering.

Are there any common mistakes to avoid when using mathematical induction?

One common mistake when using mathematical induction is assuming that the statement is true for all natural numbers without actually proving it for the base case. It is important to explicitly show that the statement is true for the first natural number. Another mistake is to assume that the statement is true for the inductive step without properly justifying it with the strong induction principle.

When should mathematical induction be used in a proof?

Mathematical induction is a useful proof technique when dealing with statements that involve natural numbers or discrete structures with a well-defined ordering. It is also helpful when the statement can be broken down into smaller cases that can be proven using the inductive step. However, it is not always the most efficient or appropriate proof technique, and other methods may be better suited for certain problems.

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