Question

Can someone please help me understand the correct steps in solving this Number Theory practice question? Thank you!

Prove that the sequence Bn we have...

n- j Bn = 1 + rt j-1 we have Bn F+1 for neN.

Answer 1

B_n = 1 + sigma^[n/2]_j = 1 (n - j i) = 1 + sigma^[n/2]_j = 1 (n - 1 - j j) + sigma^[n/2]_j = 1 (n - 1 - j j - 1) [because (r k) = (r - 1 k) + (r - 1 k - 1) = 1 + sigma^[n/2]_j = 1 (n - 1 - j j) + sigma^[n/2]_j = 1 (n - 2 - (j - 1) j - 1) n is odd therefore [n/2] = [n - 1/2] & [n - 2/2] = [n/2] - 1 therefore B_n = 1 + sigma^[n-1/2]_j = 1 (n - 1 - j j) + sigma^[n-2/2]_l = 1 (n - 2 - l l) + (n - 2 0) = B_n-1 + B_n-2 n is even therefore [n/2] = [n - 1/2] + 1, [n/2] = [n - 2 /2] + 1 therefore B_n = 1 + sigma^[n-1/2]_j = 1 (n - 1 - j j) + (n - 1 - n/2 n/2) + sigma^[n-2/2]_l = 1 (n - 2- l l) + (n - 2 0) = B_n-1 + B_n-2 B_1 = 1, B_2 = 2 & we know that F_m = F_m - 1 +