JEE Mains · Maths · STD 11 - 7. binomial theoram
Let \(\mathrm{a}=1+\frac{{ }^2 \mathrm{C}_2}{3 !}+\frac{{ }^3 \mathrm{C}_2}{4 !}+\frac{{ }^4 \mathrm{C}_2}{5 !}+\ldots\), \(\mathrm{b}=1+\frac{{ }^1 \mathrm{C}_0+{ }^1 \mathrm{C}_1}{1 !}+\frac{{ }^2 \mathrm{C}_0+{ }^2 \mathrm{C}_1+{ }^2 \mathrm{C}_2}{2 !}+\frac{{ }^3 \mathrm{C}_0+{ }^3 \mathrm{C}_1+{ }^3 \mathrm{C}_2+{ }^3 \mathrm{C}_3}{3 !}+\ldots\) Then \(\frac{2 b}{a^2}\) is equal to ...........
- A \(5\)
- B \(8\)
- C \(3\)
- D \(7\)
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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