If a1 = 2 and an = an−1 + 5, then an = 5n − 3.
any idea how i can prove this equation? much appreciated
If a1 = 2 and an = an−1 + 5, then an = 5n − 3.
any idea how i can prove this equation? much appreciated
Hello jvignacioAlthough you have headed this post 'Induction' you can prove this directly, since $\displaystyle a_1 = 2$ and $\displaystyle a_n = a_{n-1}+ 5$ is the definition of an arithmetic sequence, whose first term is 2, common difference 5. The $\displaystyle n^{th}$ term is $\displaystyle a_n=a + (n-1)d = 2 + 5(n-1) = 5n -3$
Grandad
Hello jvignacioNo, it's completely different by induction, and it looks like this:
Suppose that $\displaystyle P(n)$ is the propositional function: $\displaystyle a_n=5n-3$
Then, since $\displaystyle a_{n+1}=a_n+5, P(n)\Rightarrow a_{n+1}=5n-3+5=5(n+1)-3$
So $\displaystyle P(n) \Rightarrow P(n+1)$
Now $\displaystyle P(1)$ is: $\displaystyle a_1 = 5\times 1 -3 = 2$, which is true.
So $\displaystyle P(n)$ is true for all $\displaystyle n \ge 1$.
Do you understand what all this means?
Grandad
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):$\displaystyle a_k = 5k-3$
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
$\displaystyle P(k+1): a_{k+1} = 5(k+1)-3 $
Given
$\displaystyle a_k = 5k -3$
& $\displaystyle a_n =a_{n-1} + 5$
Proof:
$\displaystyle a_{k+1} = a_{k} + 5 $
$\displaystyle
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