Can my approach for this 'verification by Induction' be simplified?

Question

Sorry:

I don't know math scripting, so arguments maybe clumsy, sorry.

Problem (let's take sub-question (c) for simplicity)

So, the problem is:
Given: a₁ = 1 and a_n+1 = ((2n + 1)/(n+1)).a_n for all n >= 1
To find (and verify its truth by Induction): a_n

My approach:

Let P_n:
(definition for a_n I'm assuming to be true)
a_n = 1, if n = 1
OR
a_n = ((2n-1)/n).a_n-1, if n > 1

Note: The style given: a_m and a_n: a_m means value of a_m obtained by given definition (a₁ = 1 or formula for a_n+1) and a_n definition (the one I assumed to be true).

P₁:
given: a₁ = 1
a_n: a₁ = 1
given: a₁ = a_n: a₁ => P₁ is true

Now assume P_n is true:
Case 1: n = 1
a_n: a₁ = 1
a_n: a₂ = ((2(2)-1)/2).a_2-1 = 3/2
given: a₂ = ((2(1)+1)/(1+1)).a₁ = (3/2).a₁
substituting a_n: a₁ here,
given: a₂ = (3/2)
here a_n: a₂ == given: a₂
=> P₁ implies P₂ --- (1)

case 2: n > 1
a_n: a_n = ((2n-1)/n).a_n-1
a_n: a_n+1 = ((2(n+1)-1)/(n+1)).a_(n+1)-1 = ((2n+1)/(n+1)).a_n
given: a_n+1 = ((2n+1)/(n+1)).a_n
here a_n: a_n+1 = given: a_n+1
=> P_n implies P_n+1 --- (2)

(1) and (2) => P_n => P_n+1 for all n >= 1,therefore, P_n is true for all n >= 1

So, the definition of a_n is
a_n = 1, if n = 1
OR
a_n = ((2n-1)/n).a_n-1, if n > 1

question:

1. Is my approach correct?
2. Is there any simpler approach?

EDIT 1: Clarification

my question's solution is No, my approach is wrong, I need to find an explicit formula for a_n.

This may be the context you need to understand my question clearly :
My assumed formula for a_n was not an "Explicit" formula; I didn't know what it means. I transformed the given formula by replacing n with n-1 to get a formula for a_n (of the form a_n = f(a_n-1)) and thought I need to prove this transformed definition is true using induction principle, and that's what I did here. Thanks to @coffeemath for explaining what "Explicit formula" means.

Answer 1

I note you are asking only about problem 9 part (c) in which $$a_1=1,\ \ a_{n+1}=\frac{2n+1}{n+1}a_n.\tag{1}$$ I found that an explicit formula is given by $$a_n = \frac{\binom{2n}{n}}{2^n}.$$ Here the numerator is the central binomial coefficient, and the formula may be established by induction using the factorial version of the binomial coefficient, i.e. $(2n)!/(n!)^2.$ The algebra for the induction step is left for you; it is fairly straightforward using some simple properties of factorials.

added: This same recurrence and formula is associated with the Taylor series of $1/\sqrt{1-2x}$ centered at $0$ whose general term is $a_nx^{n-1}, n \ge 1.$