JEE Advance - Mathematics (2000 - No. 13)
Let $$a,\,b,\,c$$ be possitive real numbers such that $${b^2} - 4ac > 0$$ and let $${\alpha _1} = c.$$ Prove by induction that $${\alpha _{n + 1}} = {{a\alpha _n^2} \over {\left( {{b^2} - 2a\left( {{\alpha _1} + {\alpha _2} + ... + {\alpha _n}} \right)} \right)}}$$ is well-defined and
$${\alpha _{n + 1}} < {{{\alpha _n}} \over 2}$$ for all $$n = 1,2,....$$ (Here, 'well-defined' means that the denominator in the expression for $${\alpha _{n + 1}}$$ is not zero.)
$${\alpha _{n + 1}} < {{{\alpha _n}} \over 2}$$ for all $$n = 1,2,....$$ (Here, 'well-defined' means that the denominator in the expression for $${\alpha _{n + 1}}$$ is not zero.)
The problem requires proving a recursive inequality using mathematical induction.
The base case involves showing the inequality holds for n=1.
The inductive step assumes the inequality holds for n=k and proves it for n=k+1.
The condition b^2 - 4ac > 0 is crucial for ensuring the denominator is non-zero and the inequality holds.
The recursive definition of α_{n+1} involves the sum of previous α_i values.
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