Binomial Expansion 5Fe034

1. **Problem Statement:** Expand the expression $$ (1 + x)^{1/2} $$ in ascending powers of $$ x $$ up to the term involving $$ x^3 $$. 2. **Formula Used:** We use the binomial series expansion for any real exponent $$ n $$: $$ (1 + x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \cdots $$ 3. **Apply the formula:** Here, $$ n = \frac{1}{2} $$. 4. **Calculate each term:** - First term: $$ 1 $$ - Second term: $$ nx = \frac{1}{2}x $$ - Third term: $$ \frac{n(n-1)}{2!}x^2 = \frac{\frac{1}{2}(\frac{1}{2} - 1)}{2} x^2 = \frac{\frac{1}{2}(-\frac{1}{2})}{2} x^2 = \frac{-\frac{1}{4}}{2} x^2 = -\frac{1}{8} x^2 $$ - Fourth term: $$ \frac{n(n-1)(n-2)}{3!}x^3 = \frac{\frac{1}{2}(\frac{1}{2} - 1)(\frac{1}{2} - 2)}{6} x^3 $$ Calculate numerator: $$ \frac{1}{2} \times (-\frac{1}{2}) \times (-\frac{3}{2}) = \frac{1}{2} \times \frac{3}{4} = \frac{3}{8} $$ Divide by 6: $$ \frac{3}{8} \times \frac{1}{6} = \frac{3}{48} = \frac{1}{16} $$ So the fourth term is: $$ \frac{1}{16} x^3 $$ 5. **Final expansion:** $$ (1 + x)^{1/2} = 1 + \frac{1}{2}x - \frac{1}{8}x^2 + \frac{1}{16}x^3 + \cdots $$ This is the expansion up to the $$ x^3 $$ term. 6. **Explanation:** The binomial expansion allows us to write powers of binomials with fractional exponents as infinite series. Each term involves multiplying the previous term's coefficient by a factor involving $$ n $$ and decreasing integers, divided by the factorial of the term's order. This method is very useful for approximations when $$ x $$ is small.