The given expression $(1 + 2x)^{-\frac{1}{2}}$ suggests that you are looking to find the sum to infinity of a series derived from this expression. This is typically done by expanding the expression into a binomial series.

Step-by-Step Solution:

We can express $(1 + 2x)^{-\frac{1}{2}}$ as a binomial series, but this expansion is valid only when $|2x| < 1$, which implies $|x| < \frac{1}{2}$. The binomial expansion for $(1 + u)^n$, where $n = -\frac{1}{2}$ and $u = 2x$, is:

$(1 + 2x)^{-\frac{1}{2}} = 1 + \binom{-\frac{1}{2}}{1}(2x) + \binom{-\frac{1}{2}}{2}(2x)^2 + \binom{-\frac{1}{2}}{3}(2x)^3 + \dots$

The general binomial coefficient for negative powers is:

$\binom{n}{k} = \frac{n(n-1)(n-2)...(n-k+1)}{k!}$

For $n = -\frac{1}{2}$, the first few terms of the series expansion will look like:

$(1 + 2x)^{-\frac{1}{2}} = 1 - x + \frac{3}{2}x^2 - \frac{5}{2}x^3 + \dots$

Sum to Infinity

The infinite series derived from this expansion is:

$1 - x + \frac{3}{2}x^2 - \frac{5}{2}x^3 + \dots$

This is the binomial series expansion of $(1 + 2x)^{-\frac{1}{2}}$. The sum of this series continues indefinitely, but the series is convergent only for $|x| < \frac{1}{2}$.

Would you like further details or clarifications? Here are some related questions:

1. How does the binomial theorem apply to negative or fractional powers?
2. What are the conditions for the convergence of a binomial series?
3. How would you expand the expression for a different value of $n$?
4. Can you apply this approach to other functions, like $(1 - x)^{-\frac{1}{2}}$?
5. How do you compute the individual binomial coefficients for negative powers?

Tip: When dealing with binomial expansions for fractional exponents, always check the conditions for convergence, typically related to the absolute value of the variable being less than 1 for the series to converge.