Question 262 of 519MediumMCQ
If 0<x<1 and y=21x2+32x3+43x4+⋯, then the value of e1+y at x=21 is: Solutions · 1
Official solution
Writing nn−1=1−n1:
\[y = (x^2+x^3+\cdots) - \left(\frac{x^2}{2}+\frac{x^3}{3}+\cdots\right) = \frac{x^2}{1-x}+x+\ln(1-x)-x = \frac{x}{1-x}+\ln(1-x).\]
At x=21: y=1−ln2.
\[e^{1+y} = e^{2-\ln2} = \frac{e^2}{2}.\] Answer the question to unlock