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JEE Mains · Maths · STD 11 - 8. sequence and series
If \(0\,<\,x\,<\,1\) and \(y=\frac{1}{2} x^{2}+\frac{2}{3} x^{3}+\frac{3}{4} x^{4}+\ldots\), then the value of \(\mathrm{e}^{1+y}\) at \(\mathrm{x}=\frac{1}{2}\) is:
- A \(\frac{1}{2} \mathrm{e}^{2}\)
- B \(2 \mathrm{e}\)
- C \(\frac{1}{2} \sqrt{\mathrm{e}}\)
- D \(2 \mathrm{e}^{2}\)
Answer & Solution
Correct Answer
(A) \(\frac{1}{2} \mathrm{e}^{2}\)
Step-by-step Solution
Detailed explanation
\(y=\left(1-\frac{1}{2}\right) x^{2}+\left(1-\frac{1}{3}\right) x^{3}+\ldots .\) \(=\left(x^{2}+x^{3}+x^{4}+\ldots \ldots .\right)-\left(\frac{x^{2}}{2}+\frac{x^{3}}{3}+\frac{x^{4}}{4}+\ldots\right)\) \(=\frac{x^{2}}{1-x}+x-\left(x+\frac{x^{2}}{2}+\frac{x^{2}}{3}+\ldots\right)\)…
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