Below, we give a proof of the first inequality. we begin with the left side and substitute to get (k+1)a2. then, we expand the product, we substitute c1'g(k) for ka2. next, we use the inequality for the kth case. for the next step, we substitute the value of f(k) and simplify. then, we use the fact that 3ki/21 > k/2, for k greater than or equal to 2. finally, we observe that this is equal to f(k+1) to complete the proof. prove that 2(k 1) k +1 = cg(k) + 2k + 1 2 f(k) 2k+1 12 3h 12 k = f(k + 1) the second ineguality is nroven helow again we heain with the left-side suhstitute and exnand the nroduct for the next sten