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JEE Mains · Maths · STD 11 - 7. binomial theoram

ધારો કે \(\mathrm{a}=1+\frac{{ }^2 \mathrm{C}_2}{3!}+\frac{{ }^3 \mathrm{C}_2}{4!}+\frac{{ }^4 \mathrm{C}_2}{5!}+\ldots\), \(\mathrm{b}=1+\frac{{ }^1 \mathrm{C}_0+{ }^1 \mathrm{C}_1}{1!}+\frac{{ }^2 \mathrm{C}_0+{ }^2 \mathrm{C}_1+{ }^2 \mathrm{C}_2}{2!}+\frac{{ }^3 \mathrm{C}_0+{ }^3 \mathrm{C}_1+{ }^3 \mathrm{C}_2+{ }^3 \mathrm{C}_3}{3!}+\ldots .\) તો \(\frac{2 b}{a^2}=\) ...........

  1. A \(5\)
  2. B \(8\)
  3. C \(3\)
  4. D \(7\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(8\)

Step-by-step Solution

Detailed explanation

\( \mathrm{f}(\mathrm{x})=1+\frac{(1+\mathrm{x})}{1 !}+\frac{(1+\mathrm{x})^2}{2 !}+\frac{(1+\mathrm{x})^3}{3 !}+\ldots . . \)…
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