Expand (x- rac12)6 and use it to show that ( rac199400)6 ~= 0.01516.

: 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.

Step 1: Expand $(x-\frac{1}{2})^6$ using the binomial theorem.

$$(x-\frac{1}{2})^6 = \binom{6}{0}x^6(-\frac{1}{2})^0 + \binom{6}{1}x^5(-\frac{1}{2})^1 + \binom{6}{2}x^4(-\frac{1}{2})^2 + \binom{6}{3}x^3(-\frac{1}{2})^3 + \binom{6}{4}x^2(-\frac{1}{2})^4 + \binom{6}{5}x^1(-\frac{1}{2})^5 + \binom{6}{6}x^0(-\frac{1}{2})^6$$ $$= 1 \cdot x^6 \cdot 1 + 6 \cdot x^5 \cdot (-\frac{1}{2}) + 15 \cdot x^4 \cdot \frac{1}{4} + 20 \cdot x^3 \cdot (-\frac{1}{8}) + 15 \cdot x^2 \cdot \frac{1}{16} + 6 \cdot x \cdot (-\frac{1}{32}) + 1 \cdot 1 \cdot \frac{1}{64}$$ $$= x^6 - 3x^5 + \frac{15}{4}x^4 - \frac{5}{2}x^3 + \frac{15}{16}x^2 - \frac{3}{16}x + \frac{1}{64}$$

The complete expansion is $\boxed{x^6 - 3x^5 + \frac{15}{4}x^4 - \frac{5}{2}x^3 + \frac{15}{16}x^2 - \frac{3}{16}x + \frac{1}{64}}$.

Step 2: Substitute $x=\frac{1}{400}$ into the expansion.
When $x=\frac{1}{400}$, the expression $(x-\frac{1}{2})^6$ becomes $(\frac{1}{400}-\frac{1}{2})^6 = (\frac{1}{400}-\frac{200}{400})^6 = (-\frac{199}{400})^6 = (\frac{199}{400})^6$.
We evaluate the expansion with $x=\frac{1}{400} = 0.0025$. Since $x$ is small, we can use the terms in ascending powers of $x$:

$$(-\frac{1}{2}+x)^6 = \frac{1}{64} - \frac{3}{16}x + \frac{15}{16}x^2 - \frac{5}{2}x^3 + \dots$$

Substitute $x=0.0025$:

$$\frac{1}{64} = 0.015625$$ $$-\frac{3}{16}x = -\frac{3}{16}(0.0025) = -0.00046875$$ $$\frac{15}{16}x^2 = \frac{15}{16}(0.0025)^2 = \frac{15}{16}(0.00000625) = 0.000005859375$$

The terms with $x^3$ and higher powers are very small and will not affect the fifth decimal place.
Summing these values:

$$0.015625 - 0.00046875 + 0.000005859375 = 0.015162109375$$

Rounding to five decimal places, we get 0.01516.
Thus, $(\frac{199}{400})^6 \approx \boxed{0.01516}$.

: The ratio of the third to the fourth term in the expansion of $(2+3x)^n$ in ascending powers of $x$ is $5:14$ when $x=\frac{2}{5}$. Find $n$.

Step 1: Write the general term $T_{r+1}$ for $(2+3x)^n$.

$$T_{r+1} = \binom{n}{r} (2)^{n-r} (3x)^r$$

Step 2: Write the third term ($T_3$) and the fourth term ($T_4$).
For $T_3$, $r=2$:

$$T_3 = \binom{n}{2} (2)^{n-2} (3x)^2 = \binom{n}{2} 2^{n-2} 9x^2$$

For $T_4$, $r=3$:

$$T_4 = \binom{n}{3} (2)^{n-3} (3x)^3 = \binom{n}{3} 2^{n-3} 27x^3$$

Step 3: Set up the ratio $\frac{T_3}{T_4}$ and simplify.

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