Math / Algebra

QuestionTime left 0:06:46 Hide
By using proof by induction specify the firbe arrec sieps iv prove that $P(n)$ is true such that $P(n)$ is: $8+16+24+\cdots+8 n=\frac{8 n(n+1)}{2}$ is true for all positive integers $(n \geq 1)$

Studdy Solution

STEP 1

What is this asking? Prove that adding multiples of 8 up to $8n$ always equals $\frac{8n(n+1)}{2}$ for any positive integer $n$ . Watch out! Induction proofs need a strong base case and a clear inductive step!

STEP 2

1. Base Case
2. Inductive Hypothesis
3. Inductive Step

STEP 3

Let's test if the statement holds true for $n = 1$ , the smallest positive integer.
This is our base case!

STEP 4

The left side of our equation is just $8 \cdot 1 = 8$ .

STEP 5

The right side is $\frac{8 \cdot 1(1+1)}{2} = \frac{8 \cdot 2}{2} = 8$ .

STEP 6

Since both sides equal 8, the statement $P(1)$ is true!
Woohoo!

STEP 7

Let's assume the statement is true for some arbitrary positive integer $k$ .
This is our inductive hypothesis!

STEP 8

That means we're assuming $8 + 16 + 24 + \cdots + 8k = \frac{8k(k+1)}{2}$ is true.
We'll use this assumption later!

STEP 9

Now, we need to prove the statement is true for $n = k + 1$ .
This is where the real magic happens!

STEP 10

We want to show that $8 + 16 + 24 + \cdots + 8k + 8(k+1) = \frac{8(k+1)((k+1)+1)}{2}$ .
This is our target!

STEP 11

Let's start with the left side: $8 + 16 + 24 + \cdots + 8k + 8(k+1)$ .

STEP 12

Remember our inductive hypothesis?
We can replace $8 + 16 + 24 + \cdots + 8k$ with $\frac{8k(k+1)}{2}$ !
So, we have $\frac{8k(k+1)}{2} + 8(k+1)$ .

STEP 13

Let's factor out $8(k+1)$ , giving us $8(k+1)(\frac{k}{2} + 1)$ .

STEP 14

Now, let's simplify the inside of the parentheses: $\frac{k}{2} + 1 = \frac{k}{2} + \frac{2}{2} = \frac{k+2}{2}$ .

STEP 15

Putting it all together, we get $8(k+1)(\frac{k+2}{2}) = \frac{8(k+1)(k+2)}{2}$ .

STEP 16

Notice that $k+2$ is the same as $(k+1) + 1$ .
So, we have $\frac{8(k+1)((k+1)+1)}{2}$ , which is exactly what we wanted to show!
Boom!

STEP 17

By the principle of mathematical induction, we have proven that $8 + 16 + 24 + \cdots + 8n = \frac{8n(n+1)}{2}$ is true for all positive integers $n$ .

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QuestionTime left 0:06:46 Hide
By using proof by induction specify the firbe arrec sieps iv prove that $P(n)$ is true such that $P(n)$ is: $8+16+24+\cdots+8 n=\frac{8 n(n+1)}{2}$ is true for all positive integers $(n \geq 1)$

Studdy Solution

STEP 1

What is this asking? Prove that adding multiples of 8 up to $8n$ always equals $\frac{8n(n+1)}{2}$ for any positive integer $n$ . Watch out! Induction proofs need a strong base case and a clear inductive step!

STEP 2

1. Base Case
2. Inductive Hypothesis
3. Inductive Step

STEP 3

Let's test if the statement holds true for $n = 1$ , the smallest positive integer.
This is our base case!

STEP 4

The left side of our equation is just $8 \cdot 1 = 8$ .

STEP 5

The right side is $\frac{8 \cdot 1(1+1)}{2} = \frac{8 \cdot 2}{2} = 8$ .

STEP 6

Since both sides equal 8, the statement $P(1)$ is true!
Woohoo!

STEP 7

Let's assume the statement is true for some arbitrary positive integer $k$ .
This is our inductive hypothesis!

STEP 8

That means we're assuming $8 + 16 + 24 + \cdots + 8k = \frac{8k(k+1)}{2}$ is true.
We'll use this assumption later!

STEP 9

Now, we need to prove the statement is true for $n = k + 1$ .
This is where the real magic happens!

STEP 10

We want to show that $8 + 16 + 24 + \cdots + 8k + 8(k+1) = \frac{8(k+1)((k+1)+1)}{2}$ .
This is our target!

STEP 11

Let's start with the left side: $8 + 16 + 24 + \cdots + 8k + 8(k+1)$ .

STEP 12

Remember our inductive hypothesis?
We can replace $8 + 16 + 24 + \cdots + 8k$ with $\frac{8k(k+1)}{2}$ !
So, we have $\frac{8k(k+1)}{2} + 8(k+1)$ .

STEP 13

Let's factor out $8(k+1)$ , giving us $8(k+1)(\frac{k}{2} + 1)$ .

STEP 14

Now, let's simplify the inside of the parentheses: $\frac{k}{2} + 1 = \frac{k}{2} + \frac{2}{2} = \frac{k+2}{2}$ .

STEP 15

Putting it all together, we get $8(k+1)(\frac{k+2}{2}) = \frac{8(k+1)(k+2)}{2}$ .

STEP 16

Notice that $k+2$ is the same as $(k+1) + 1$ .
So, we have $\frac{8(k+1)((k+1)+1)}{2}$ , which is exactly what we wanted to show!
Boom!

STEP 17

By the principle of mathematical induction, we have proven that $8 + 16 + 24 + \cdots + 8n = \frac{8n(n+1)}{2}$ is true for all positive integers $n$ .

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QuestionTime left 0:06:46 HideBy using proof by induction specify the firbe arrec sieps iv prove that P(n)P(n)P(n) is true such that P(n)P(n)P(n) is: 8+16+24+⋯+8n=8n(n+1)28+16+24+\cdots+8 n=\frac{8 n(n+1)}{2}8+16+24+⋯+8n=28n(n+1)​ is true for all positive integers (n≥1)(n \geq 1)(n≥1)

Studdy Solution

STEP 1

What is this asking? Prove that adding multiples of 8 up to 8n8n8n always equals 8n(n+1)2\frac{8n(n+1)}{2}28n(n+1)​ for any positive integer nnn. Watch out! Induction proofs need a strong base case and a clear inductive step!

STEP 2

1. Base Case2. Inductive Hypothesis3. Inductive Step

STEP 3

Let's **test** if the statement holds true for n=1n = 1n=1, the smallest positive integer.This is our **base case**!

STEP 4

The left side of our equation is just 8⋅1=88 \cdot 1 = 88⋅1=8.

STEP 5

The right side is 8⋅1(1+1)2=8⋅22=8\frac{8 \cdot 1(1+1)}{2} = \frac{8 \cdot 2}{2} = 828⋅1(1+1)​=28⋅2​=8.

STEP 6

Since both sides equal **8**, the statement P(1)P(1)P(1) is **true**!Woohoo!

STEP 7

Let's **assume** the statement is true for some arbitrary positive integer kkk.This is our **inductive hypothesis**!

STEP 8

That means we're assuming 8+16+24+⋯+8k=8k(k+1)28 + 16 + 24 + \cdots + 8k = \frac{8k(k+1)}{2}8+16+24+⋯+8k=28k(k+1)​ is **true**.We'll use this assumption later!

STEP 9

Now, we need to **prove** the statement is true for n=k+1n = k + 1n=k+1.This is where the real magic happens!

STEP 10

We want to show that 8+16+24+⋯+8k+8(k+1)=8(k+1)((k+1)+1)28 + 16 + 24 + \cdots + 8k + 8(k+1) = \frac{8(k+1)((k+1)+1)}{2}8+16+24+⋯+8k+8(k+1)=28(k+1)((k+1)+1)​.This is our **target**!

STEP 11

Let's start with the left side: 8+16+24+⋯+8k+8(k+1)8 + 16 + 24 + \cdots + 8k + 8(k+1)8+16+24+⋯+8k+8(k+1).

STEP 12

Remember our **inductive hypothesis**?We can replace 8+16+24+⋯+8k8 + 16 + 24 + \cdots + 8k8+16+24+⋯+8k with 8k(k+1)2\frac{8k(k+1)}{2}28k(k+1)​!So, we have 8k(k+1)2+8(k+1)\frac{8k(k+1)}{2} + 8(k+1)28k(k+1)​+8(k+1).

STEP 13

Let's **factor out** 8(k+1)8(k+1)8(k+1), giving us 8(k+1)(k2+1)8(k+1)(\frac{k}{2} + 1)8(k+1)(2k​+1).

STEP 14

Now, let's simplify the inside of the parentheses: k2+1=k2+22=k+22\frac{k}{2} + 1 = \frac{k}{2} + \frac{2}{2} = \frac{k+2}{2}2k​+1=2k​+22​=2k+2​.

STEP 15

Putting it all together, we get 8(k+1)(k+22)=8(k+1)(k+2)28(k+1)(\frac{k+2}{2}) = \frac{8(k+1)(k+2)}{2}8(k+1)(2k+2​)=28(k+1)(k+2)​.

STEP 16

Notice that k+2k+2k+2 is the same as (k+1)+1(k+1) + 1(k+1)+1.So, we have 8(k+1)((k+1)+1)2\frac{8(k+1)((k+1)+1)}{2}28(k+1)((k+1)+1)​, which is exactly what we wanted to show!Boom!

STEP 17

By the principle of mathematical induction, we have proven that 8+16+24+⋯+8n=8n(n+1)28 + 16 + 24 + \cdots + 8n = \frac{8n(n+1)}{2}8+16+24+⋯+8n=28n(n+1)​ is true for all positive integers nnn.

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QuestionTime left 0:06:46 Hide
By using proof by induction specify the firbe arrec sieps iv prove that $P(n)$ is true such that $P(n)$ is: $8+16+24+\cdots+8 n=\frac{8 n(n+1)}{2}$ is true for all positive integers $(n \geq 1)$

What is this asking? Prove that adding multiples of 8 up to $8n$ always equals $\frac{8n(n+1)}{2}$ for any positive integer $n$ . Watch out! Induction proofs need a strong base case and a clear inductive step!

1. Base Case
2. Inductive Hypothesis
3. Inductive Step

Let's test if the statement holds true for $n = 1$ , the smallest positive integer.
This is our base case!

The left side of our equation is just $8 \cdot 1 = 8$ .

The right side is $\frac{8 \cdot 1(1+1)}{2} = \frac{8 \cdot 2}{2} = 8$ .

Since both sides equal 8, the statement $P(1)$ is true!
Woohoo!

Let's assume the statement is true for some arbitrary positive integer $k$ .
This is our inductive hypothesis!

That means we're assuming $8 + 16 + 24 + \cdots + 8k = \frac{8k(k+1)}{2}$ is true.
We'll use this assumption later!

Now, we need to prove the statement is true for $n = k + 1$ .
This is where the real magic happens!

We want to show that $8 + 16 + 24 + \cdots + 8k + 8(k+1) = \frac{8(k+1)((k+1)+1)}{2}$ .
This is our target!

Let's start with the left side: $8 + 16 + 24 + \cdots + 8k + 8(k+1)$ .

Remember our inductive hypothesis?
We can replace $8 + 16 + 24 + \cdots + 8k$ with $\frac{8k(k+1)}{2}$ !
So, we have $\frac{8k(k+1)}{2} + 8(k+1)$ .

Let's factor out $8(k+1)$ , giving us $8(k+1)(\frac{k}{2} + 1)$ .

Now, let's simplify the inside of the parentheses: $\frac{k}{2} + 1 = \frac{k}{2} + \frac{2}{2} = \frac{k+2}{2}$ .

Putting it all together, we get $8(k+1)(\frac{k+2}{2}) = \frac{8(k+1)(k+2)}{2}$ .

Notice that $k+2$ is the same as $(k+1) + 1$ .
So, we have $\frac{8(k+1)((k+1)+1)}{2}$ , which is exactly what we wanted to show!
Boom!

By the principle of mathematical induction, we have proven that $8 + 16 + 24 + \cdots + 8n = \frac{8n(n+1)}{2}$ is true for all positive integers $n$ .