Answer Happy

Question 2: Define two sequences, ai and bi by an = 4(n+3)(n+2)(n+1)(n) and bn = 120 (n+4)(n+3)(n+2)(n+1)(n). (a) Compute ao, a1, a2, a3, a4, and a5. (b) Compute bo, b1, b2, b3, b4, and b5. (c) Check that b5 - b4 = a5. (d) Use algebra to show that bn -bn-1 = an for every n ≥ 1. Hint: It is not necessary to expand these polynomials, since they have many common factors. k (e) Use the previous part to conclude that Σai = [(bi-bi-1). i=1 i=1 k (f) The right hand sum is called a telescoping sum, is equal to bk - bo, since all other terms cancel. Check your work by computing a₁ + a₂ + a3 + a₁ + a5 and comparing with b5. Note that both the ai's and the b; occur as entries along diagonals of Pascal's triangle. It is relatively easy to check that n = an+11an-1+11an-2+an-3 and we could use this to find a formula for 14+24¹ +3¹ +...+nª.