1. ## Induction question help

If a1 = 2 and an = an−1 + 5, then an = 5n − 3.

any idea how i can prove this equation? much appreciated

2. ## Sequence

Hello jvignacio
Originally Posted by jvignacio
If a1 = 2 and an = an1 + 5, then an = 5n 3.

any idea how i can prove this equation? much appreciated
Although you have headed this post 'Induction' you can prove this directly, since $a_1 = 2$ and $a_n = a_{n-1}+ 5$ is the definition of an arithmetic sequence, whose first term is 2, common difference 5. The $n^{th}$ term is $a_n=a + (n-1)d = 2 + 5(n-1) = 5n -3$

Hello jvignacioAlthough you have headed this post 'Induction' you can prove this directly, since $a_1 = 2$ and $a_n = a_{n-1}+ 5$ is the definition of an arithmetic sequence, whose first term is 2, common difference 5. The $n^{th}$ term is $a_n=a + (n-1)d = 2 + 5(n-1) = 5n -3$

ahhh yes thank you !

Hello jvignacioAlthough you have headed this post 'Induction' you can prove this directly, since $a_1 = 2$ and $a_n = a_{n-1}+ 5$ is the definition of an arithmetic sequence, whose first term is 2, common difference 5. The $n^{th}$ term is $a_n=a + (n-1)d = 2 + 5(n-1) = 5n -3$

Hi Grandad, if we were to prove this by induction proof, would it be done the same way ? because i didnt see you tryed the base case first then n+1. Cheers

5. ## Induction proof

Hello jvignacio
Originally Posted by jvignacio
Hi Grandad, if we were to prove this by induction proof, would it be done the same way ? because i didnt see you tryed the base case first then n+1. Cheers
No, it's completely different by induction, and it looks like this:

Suppose that $P(n)$ is the propositional function: $a_n=5n-3$

Then, since $a_{n+1}=a_n+5, P(n)\Rightarrow a_{n+1}=5n-3+5=5(n+1)-3$

So $P(n) \Rightarrow P(n+1)$

Now $P(1)$ is: $a_1 = 5\times 1 -3 = 2$, which is true.

So $P(n)$ is true for all $n \ge 1$.

Do you understand what all this means?

Hello jvignacioNo, it's completely different by induction, and it looks like this:

Suppose that $P(n)$ is the propositional function: $a_n=5n-3$

Then, since $a_{n+1}=a_n+5, P(n)\Rightarrow a_{n+1}=5n-3+5=5(n+1)-3$

So $P(n) \Rightarrow P(n+1)$

Now $P(1)$ is: $a_1 = 5\times 1 -3 = 2$, which is true.

So $P(n)$ is true for all $n \ge 1$.

Do you understand what all this means?

hrmm not really. you lost me at the beginning.. i will try study it more

7. Originally Posted by jvignacio
If a1 = 2 and an = an−1 + 5, then an = 5n − 3.

any idea how i can prove this equation? much appreciated
Proof by induction

Lets consider this as a statement
(this means P(h) addresses for n= h)

Show it for n= 1

P(1): a1= 5(1) -3 = 2

Its given in the question that a1 =2
Thus this statement is correct for n=1

Assume it correct for n= k

So we have assumed that
P(k) is correct this means

P(k): $a_k = 5k-3$

Using above assumption prove it for n= k+1

So we need to prove that P(k+1) is correct using the assumption that P(k) is correct
steps
TO Prove
$P(k+1): a_{k+1} = 5(k+1)-3$

Given
$a_k = 5k -3$

& $a_n =a_{n-1} + 5$

Proof:

$a_{k+1} = a_{k} + 5$

$
a_{k+1} = 5k-3 + 5 = 5(k+1) -3$

This proves that its correct for k+1

Hence its correct for all real numbers through the principle of mathematical induction

Proof by induction

Lets consider this as a statement
(this means P(h) addresses for n= h)

Show it for n= 1

P(1): a1= 5(1) -3 = 2

Its given in the question that a1 =2
Thus this statement is correct for n=1

Assume it correct for n= k

So we have assumed that
P(k) is correct this means

P(k): $a_k = 5k-3$

Using above assumption prove it for n= k+1

So we need to prove that P(k+1) is correct using the assumption that P(k) is correct
steps
TO Prove
$P(k+1): a_{k+1} = 5(k+1)-3$

Given
$a_k = 5k -3$

& $a_n =a_{n-1} + 5$

Proof:

$a_{k+1} = a_{k} + 5$

$
a_{k+1} = 5k-3 + 5 = 5(k+1) -3$

This proves that its correct for k+1

Hence its correct for all real numbers through the principle of mathematical induction
thanks for that. really cleared things up well for me.

jv