Let a and b be positive real numbers, with a>b. Compute ba1+a(2a−b)1+(2a−b)(3a−2b)1+(3a−2b)(4a−3b)1+⋯.
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The nth term is [(n−1)a−(n−2)b][na−(n−1)b]1.We can write [(n−1)a−(n−2)b][na−(n−1)b]1=(a−b)[(n−1)a−(n−2)b][na−(n−1)b]a−b=(a−b)[(n−1)a−(n−2)b][na−(n−1)b][na−(n−1)b]−[(n−1)a−(n−2)b]=(a−b)[(n−1)a−(n−2)b]1−(a−b)[na−(n−1)b]1.Thus, ba1+a(2a−b)1+(2a−b)(3a−2b)1+(3a−2b)(4a−3b)1+⋯=((a−b)b1−(a−b)a1)+((a−b)a1−(a−b)(2a−b)1)+((a−b)(2a−b)1−(a−b)(3a−2b)1)+⋯=(a−b)b1.