You need to evaluate `I_n` using integration by parts, such that:

`I_n = int_0^1 x^n*sqrt(1 - x)dx`

`u = x^n => du = n*x^(n-1)`

`dv = sqrt(1 - x)dx => v = int sqrt(1 - x)dx = -(2/3)(1-x)^(3/2)`

Using the formula of integration by parts yields:

`int udv = uv - int vdu`

Reasoning by analogy yields:

`I_n = -(2/3)x^n*(1-x)^(3/2)|_0^1 + (2n)/3 int_0^1 x^(n-1)(1 - x)sqrt(1 - x)dx`

Using the fundamental theorem of calculus yields:

`I_n = -(2/3)1^n*(1-1)^(3/2) + (2/3)0^n*(1-0)^(3/2) + (2n)/3 int_0^1 (x^(n-1))sqrt(1 - x) - x^n*sqrt(1 - x))dx`

Using the property of linearity of integral yields:

`I_n = (2n)/3 int_0^1 (x^(n-1))sqrt(1 - x))dx - (2n)/3 int_0^1 (x^n*sqrt(1 - x))dx`

Since `I_n = int_0^1 (x^n*sqrt(1 - x))dx` , hence, by analogy, `I_(n-1) = int_0^1 (x^(n-1))sqrt(1 - x))dx` .

`I_n = (2n)/3 I_(n-1) - (2n)/3 I_n`

Moving the term that contain `I_n` to the left side, yields:

`I_n + (2n)/3 I_n = (2n)/3 I_(n-1)`

Factoring out `I_n` yields:

`I_n(1 + (2n)/3) = (2n)/3 I_(n-1)`

`I_n (3 + 2n)/3 = (2n)/3 I_(n-1) => I_n = ((2n)/3)/((3 + 2n)/3) I_(n-1) `

Reducing duplicate factors yields:

`I_n = (2n)/(3 + 2n) I_(n-1) `

**Hence, testing if `I_n = (2n)/(3 + 2n) I_(n-1)` holds, under the given conditions, the answer is affirmative.**