f6 pure maths(急!!!!!)

1) When n=0, 2 * (0+2) a2 - 3 * 0 * a1 + ( 0 - 1) a0 = 0
=> 4a2 - a0 = 0 => 4a2 - 1 = 0 => a2 = 1/4 = 1/2^2

When n=1, 2*(1+2) a3 - 3 * 1 * a2 + (1-1) a1 = 0
=> 6 a3 - 3 a2 = 0 => 6a3 - 3 * 1/ 4=0 => a3 = 1/8 = 1/2^3

If hypothesis holds for k and k + 1, when n = k
2 * ( k + 2 ) a(k+2) - 3 * k * a(k+1) + (k - 1)ak = 0
2 * ( k + 2 ) a(k+2) - 3 * k * 1/2^(k+1) + (k - 1) * 1 / 2^k = 0 (induction hypothesis)
2 * ( k + 2 ) a(k+2) = (k + 2) / 2^(k+1)
a(k+2) = 1 / 2^(k+2)
So hypothesis holds for k+2 if it holds for k and k+1
By Mathematical Induction, the hypothesis holds for all integers >= 2.


2) a1 = 1 = 1/1^2
If hypothesis holds for k, when n = k,
1/a(k+1) - 1/ak = 2k+1
1/a(k+1) - k^2 = 2k + 1 (induction hypothesis)
1/a(k+1) = k^2 + 2k + 1 = (k+1)^2
=> a(k+1) = 1/ (k+1)^2
So hypothesis holds for k+1 if it holds for k
By Mathematical Induction, the hypothesis holds for all positive integers.