Question 2: 1 Define two sequences, a; and b; by an = (n+3)(n+2)(n+1)(n) and bn = (n+4)(n+3)(n+2)(n+1)(n). 120 (a) Compu

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Question 2: 1 Define two sequences, a; and b; by an = (n+3)(n+2)(n+1)(n) and bn = (n+4)(n+3)(n+2)(n+1)(n). 120 (a) Compu

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Question 2 1 Define Two Sequences A And B By An N 3 N 2 N 1 N And Bn N 4 N 3 N 2 N 1 N 120 A Compu 1
Question 2 1 Define Two Sequences A And B By An N 3 N 2 N 1 N And Bn N 4 N 3 N 2 N 1 N 120 A Compu 1 (197.38 KiB) Viewed 13 times
Question 2: 1 Define two sequences, a; and b; by an = (n+3)(n+2)(n+1)(n) and bn = (n+4)(n+3)(n+2)(n+1)(n). 120 (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 Σ Ai = i=1 (bi-bi-1). (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. i=1 Note that both the ai's and the bi occur as entries along diagonals of Pascal's triangle. It is relatively easy to check that n¹ = an+11an-1+11an-2+an-3 and we could use this to find a formula for 14+24+34 +...+nª.
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