7. Write down in its simplest form the complete expansion of $(x-\frac{1}{2})^6$. Hence, by taking $x=\frac{1}{400}$ in this expansion, show that $(\frac{199}{400})^6 = 0.01516$, correct to five places of decimals. Part 1: Complete expansion of $(x-\frac{1}{2})^6$. Using the binomial theorem $(A+B)^n = \sum_{k=0}^{n} \binom{n}{k} A^{n-k} B^k$ with $A=x$, $B=-\frac{1}{2}$, and $n=6$: $$ \left(x-\frac{1}{2}\right)^6 = \binom{6}{0}x^6\left(-\frac{1}{2}\right)^0 + \binom{6}{1}x^5\left(-\frac{1}{2}\right)^1 + \binom{6}{2}x^4\left(-\frac{1}{2}\right)^2 + \binom{6}{3}x^3\left(-\frac{1}{2}\right)^3 + \binom{6}{4}x^2\left(-\frac{1}{2}\right)^4 + \binom{6}{5}x^1\left(-\frac{1}{2}\right)^5 + \binom{6}{6}x^0\left(-\frac{1}{2}\right)^6 $$ Calculate the binomial coefficients and powers of $-\frac{1}{2}$: $$ = 1 \cdot x^6 \cdot 1 + 6 \cdot x^5 \cdot \left(-\frac{1}{2}\right) + 15 \cdot x^4 \cdot \left(\frac{1}{4}\right) + 20 \cdot x^3 \cdot \left(-\frac{1}{8}\right) + 15 \cdot x^2 \cdot \left(\frac{1}{16}\right) + 6 \cdot x^1 \cdot \left(-\frac{1}{32}\right) + 1 \cdot x^0 \cdot \left(\frac{1}{64}\right) $$ Simplify each term: $$ = x^6 - 3x^5 + \frac{15}{4}x^4 - \frac{5}{2}x^3 + \frac{15}{16}x^2 - \frac{3}{16}x + \frac{1}{64} $$ The complete expansion is $\boxed{x^6 - 3x^5 + \frac{15}{4}x^4 - \frac{5}{2}x^3 + \frac{15}{16}x^2 - \frac{3}{16}x + \frac{1}{64}}$. Part 2: Show that $(\frac{199}{400})^6 = 0.01516$ by taking $x=\frac{1}{400}$. Substitute $x=\frac{1}{400}$ into the expression $(x-\frac{1}{2})^6$: $$ \left(\frac{1}{400}-\frac{1}{2}\right)^6 = \left(\frac{1}{400}-\frac{200}{400}\right)^6 = \left(-\frac{199}{400}\right)^6 = \left(\frac{199}{400}\right)^6 $$ Now substitute $x=\frac{1}{400}$ into the expanded form: $$ \left(\frac{1}{400}\right)^6 - 3\left(\frac{1}{400}\right)^5 + \frac{15}{4}\left(\frac{1}{400}\right)^4 - \frac{5}{2}\left(\frac{1}{400}\right)^3 + \frac{15}{16}\left(\frac{1}{400}\right)^2 - \frac{3}{16}\left(\frac{1}{400}\right) + \frac{1}{64} $$ Evaluate each term: $$ \left(\frac{1}{400}\right)^6 \approx 2.44 \times 10^{-16} $$ $$ -3\left(\frac{1}{400}\right)^5 \approx -2.86 \times 10^{-13} $$ $$ \frac{15}{4}\left(\frac{1}{400}\right)^4 \approx 1.46 \times 10^{-10} $$ $$ -\frac{5}{2}\left(\frac{1}{400}\right)^3 \approx -3.91 \times 10^{-8} $$ $$ \frac{15}{16}\left(\frac{1}{400}\right)^2 = \frac{15}{16} \cdot \frac{1}{16000