#1537 · Algebra · Sequences & series

Alternating Sum of q Products

If $q_{n} = \frac{1 \cdot 3 \cdot 5 \cdot 7 \dots ( 2 n - 1 )}{2 \cdot 4 \cdot 6 \cdot 8 \dots 2 n}$ , prove that $2 {q_{2 n} - q_{1} q_{2 n - 1} + q_{2} q_{2 n - 2} + \dots + (- 1)^{n - 1} q_{n - 1} q_{n + 1}} = q_{n} + (- 1)^{n - 1} q_{n}^{2}$ .

~30 minProof · checked by AIBinomial TheoremGenerating Functions

H. S. Hall & S. R. Knight, Higher Algebra (1891) · Ch. XIV. Binomial Theorem: Any Index, Examples XIV.c, No. 20 (part 2)

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