If 0 \(<\) x \(<\) 1 and y = \(\frac{1}{2}x^2 + \frac{2}{3}x^3 + \frac{3}{4}x^4 + ...\), then the value of e\(^{1+y}\) at x = \(\frac{1}{2}\) is :
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When faced with an infinite series, try to break it down into combinations of standard Maclaurin series like the geometric series \(\frac{1}{1-x}\) and the logarithmic series \(\ln(1-x)\) or \(\ln(1+x)\). Differentiating or integrating the series can sometimes help in recognizing the pattern.
Step 1: Express the series in a recognizable form.
The given series for y is:
\[ y = \frac{1}{2}x^2 + \frac{2}{3}x^3 + \frac{3}{4}x^4 + \dots \]
The general term can be written as \(\frac{n-1}{n}x^n\). We can split this term:
\[ y = \sum_{n=2}^{\infty} \left(1 - \frac{1}{n}\right)x^n = \sum_{n=2}^{\infty} x^n - \sum_{n=2}^{\infty} \frac{x^n}{n} \]
Step 2: Evaluate the two resulting series.
The first series is a standard geometric series starting from \(x^2\):
\[ \sum_{n=2}^{\infty} x^n = x^2 + x^3 + x^4 + \dots = \frac{\text{first term}}{1 - \text{common ratio}} = \frac{x^2}{1-x} \]
The second series is related to the Maclaurin series for \(\ln(1-x)\):
\[ \ln(1-x) = -x - \frac{x^2}{2} - \frac{x^3}{3} - \frac{x^4}{4} - \dots \]
From this, we can see that:
\[ \sum_{n=2}^{\infty} \frac{x^n}{n} = \frac{x^2}{2} + \frac{x^3}{3} + \dots = -\ln(1-x) - x \]
Step 3: Combine the expressions to find y.
\[ y = \left(\frac{x^2}{1-x}\right) - \left(-\ln(1-x) - x\right) = \frac{x^2}{1-x} + x + \ln(1-x) \]
Taking a common denominator for the first two terms:
\[ y = \frac{x^2 + x(1-x)}{1-x} + \ln(1-x) = \frac{x^2 + x - x^2}{1-x} + \ln(1-x) \]
\[ y = \frac{x}{1-x} + \ln(1-x) \]
Step 4: Find the expression for e\(^{1+y}\).
First, find \(1+y\):
\[ 1+y = 1 + \frac{x}{1-x} + \ln(1-x) = \frac{1-x+x}{1-x} + \ln(1-x) = \frac{1}{1-x} + \ln(1-x) \]
Now, exponentiate this expression:
\[ e^{1+y} = e^{\frac{1}{1-x} + \ln(1-x)} = e^{\frac{1}{1-x}} \cdot e^{\ln(1-x)} = (1-x)e^{\frac{1}{1-x}} \]
Step 5: Substitute x = 1/2.
\[ e^{1+y} = \left(1 - \frac{1}{2}\right) e^{\frac{1}{1 - 1/2}} = \left(\frac{1}{2}\right) e^{\frac{1}{1/2}} = \frac{1}{2} e^2 \]
