Question

3. The sequence bois defined as follows: boo, and for integers n 2 2, bn V1 (a) Calculate ba, ba, b4 and bs. (b) Use part (a) to guess a formula for bn for all integers n 2 0. (c) Prove by induction on n that your guess in part (b) is correct. Reflect in ePerttolio Downloard MacBook Air 80

Answer 1

(a) For n=2, b₂ =

Vbibo +3(2)/2 -1

V(1/2)0+3-1

=>

2

For n =3,

ba = Vb2b1 + 3(3)/2-1

=> $b_3=sqrt{2(1/2)}+9/2-1$

=> $b_3=1+9/2-1=9/2$

For n=4., $b_4=sqrt{b_3b_2}+3(4)/2-1$

=> $b_4=sqrt{2(9/2)}+6-1$

=> $b_4=sqrt{(9)}+6-1=3+5=8$

For n = 5, $b_5=sqrt{b_4b_3}+3(5)/2-1$

=> $b_5=sqrt{8(9/2)}+15/2-1$

=>

bs = V36 + 13/2

=> $b_5=6+13/2=25/2$

b₂ = 2, b₃ = 9/2, b₄ = 8, b₅ = 25/2

(b) Looking at the pattern,

b_(2n) = 2n², b_(2n+1) = (2n+1)²/2

(c) For n = 0, b₀ = 0 , b₁ = 1/2

Suppose the result is true for all values less than n

For n+1, b_2(n+1) = b_2n+2 =

Vb2n+1b2n +3(2n 2)/2 -1

=> $b_{2n+2}=sqrt{(2n+1)^2/2(2n^2)}+3*2(n+1)/2-1$

=>

b2n +2 = V/(2n + 1)2(n*) + 3(n + 1)-1

=> $b_{2n+2}={(2n+1)(n)}+3(n+1)-1$

=>

2n+2

=> $b_{2n+2}=2n^2+4n+2$

=>

n-+2

=>

62n+2 2(n +1)2

Similarly b_2n = 2n²

Hence our guess is true