Please I need help with this induction problems as soon as possible. Please helpas soon as possible.?

5.
3^1 > 1^2
So, it is true for n = 1

Suppose it is true for n = k , where k ≧ 2
That is , 3^k > k^2 ..... (1)

When n = k+1 ,
3^(k+1) - (k+1)^2
= 3*(3^k) - (k+1)^2
> 3*(k^2) - (k+1)^2 , by (1)
= (√3 k)^2 - (k+1)^2
= (√3 k + k + 1 )(√3 k - k - 1 )
= [ (√3 + 1)k + 1 ] * [ (√3 - 1)k - 1 ]

Since k ≧ 2 ,
(√3 + 1)k + 1 > 0

(√3 - 1)k - 1 ≧ (√3 - 1)*2 - 1 = 2√3 - 3 = √12 - √9 > 0

So, [ (√3 + 1)k + 1 ] * [ (√3 - 1)k - 1 ] > 0
Hence, 3^(k+1) - (k+1)^2 > 0
3^(k+1) > (k+1)^2
So, it is true for n = k+1.

By the mathematical induction, it is true for all positive integers n.
Q.E.D.

6.
0 < X0 = 0.5 < 1
x1 = 0.5*x0*( 1 - x0 ) = 0.5*0.5*( 1 - 0.5 ) = (1/2)^3 = 1/8
So, 0 < x1 = 1/8 < 1
Thus, it is true for n = 0 and n = 1.

Suppose it is true for n = k , where k ≧ 2
That is , 0 < xk < 1 ..... (1)

So,
- 1 < - xk < 0
0 < 1 - xk < 1 ..... (2)
0 < xk*( 1 - xk ) < 1 , by (1) & (2)
0 < 0.5*xk*( 1 - xk ) < 0.5
0 < xk+1 < 0.5 < 1
So, it is true for n = k+1

By the mathematical induction, it is true for all non-negative integers n.
Q.E.D.