[JEE Advanced 2001] let a b c 0 with b 2 4ac 0 and alpha 1 c
Let $a,b,c>0$ with $b^2-4ac>0$ and $\alpha_1=c$. Prove by induction that $\alpha_{n+1}=\dfrac{a\alpha_n^2}{b^2-2a(\alpha_1+\cdots+\alpha_n)}$ is well-defined and $\alpha_{n+1}<\dfrac{\alpha_n}2$.
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✓ Vidaara Team
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· 2d ago
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Answer: Proved.
Induction shows the denominator stays positive (bounded below by $b^2-4ac>0$) and that $\frac{a\alpha_n}{b^2-2a\sum\alpha_i}<\frac12$, giving $\alpha_{n+1}<\frac{\alpha_n}2$.
JEE Advanced 2001 · Binomial Theorem — verified solution by the Vidaara Team.
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