Question
1. Without using table, find the value of
(2+root3)^6+(2-root3)^6
2. Using the binomial theorem to evaluate 1.0048 correct to 5d.p
(2+root3)^6+(2-root3)^6
2. Using the binomial theorem to evaluate 1.0048 correct to 5d.p
Answers
GPT-4o mini
### Problem 1:
We want to calculate \( (2 + \sqrt{3})^6 + (2 - \sqrt{3})^6 \).
Let \( a = 2 + \sqrt{3} \) and \( b = 2 - \sqrt{3} \). Notice that \( a \) and \( b \) are conjugates, and their product is:
\[
ab = (2 + \sqrt{3})(2 - \sqrt{3}) = 4 - 3 = 1
\]
This indicates that \( b = \frac{1}{a} \). Therefore, we can express \( b^6 = \left(\frac{1}{a}\right)^6 = \frac{1}{a^6} \).
Now, we want to find:
\[
a^6 + b^6 = a^6 + \frac{1}{a^6}
\]
Next, we can express \( a^6 + \frac{1}{a^6} \) using the identity:
\[
x^6 + \frac{1}{x^6} = \left(x^3 + \frac{1}{x^3}\right)^2 - 2
\]
First, we need to find \( a + b \) and \( ab \).
\[
a + b = (2 + \sqrt{3}) + (2 - \sqrt{3}) = 4
\]
\[
ab = 1
\]
Next, we can find \( a^n + b^n \) using the recurrence relation:
\[
s_n = a^n + b^n
\]
with initial conditions \( s_0 = 2 \), \( s_1 = 4 \) and the recurrence relation:
\[
s_n = (a + b)s_{n-1} - ab s_{n-2} = 4s_{n-1} - s_{n-2}
\]
Now we calculate \( s_n \):
- \( s_2 = 4s_1 - s_0 = 4 \cdot 4 - 2 = 16 - 2 = 14 \)
- \( s_3 = 4s_2 - s_1 = 4 \cdot 14 - 4 = 56 - 4 = 52 \)
- \( s_4 = 4s_3 - s_2 = 4 \cdot 52 - 14 = 208 - 14 = 194 \)
- \( s_5 = 4s_4 - s_3 = 4 \cdot 194 - 52 = 776 - 52 = 724 \)
- \( s_6 = 4s_5 - s_4 = 4 \cdot 724 - 194 = 2896 - 194 = 2702 \)
Thus, the value of \( (2+\sqrt{3})^6 + (2-\sqrt{3})^6 \) is:
\[
\boxed{2702}
\]
---
### Problem 2:
We need to evaluate \( 1.0048 \) using the binomial theorem, writing it as \( 1 + 0.0048 \), and apply the binomial expansion for \( (1 + x)^n \).
Using the binomial expansion,
\[
(1 + x)^n = \sum_{k=0}^{n} \binom{n}{k} x^k
\]
We choose \( n = 1 \) and \( x = 0.0048 \):
\[
(1.0048)^{1} = 1 + \binom{1}{1} (0.0048) = 1 + 0.0048 = 1.0048
\]
However, let's consider \( n = \frac{1}{2} \) for demonstration. We'll try evaluating it with \( n = 1 + \frac{1}{4} \) to find an approximation:
To expand to five decimal places and for greater accuracy, using up to the \( x^4 \) term:
\[
(1 + x)^{n} \approx 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \frac{n(n-1)(n-2)(n-3)}{4!}x^4
\]
Let \( n = 1 \) and \( x = 0.0048 \):
1. \( 1 \)
2. \( nx = 1 \times 0.0048 = 0.0048 \)
3. \( \frac{1 \cdot 0}{2} \cdot (0.0048)^2 = 0 \)
4. \( \frac{1 \cdot 0 \cdot -1}{3} \cdot (0.0048)^3 = 0 \)
5. \( \frac{1 \cdot 0 \cdot -1 \cdot -2}{4!} \cdot (0.0048)^4 = 0 \)
Taking the above terms into account:
Thus the result is:
\[
1.0048
\]
For more precision, stop here; otherwise, numerical methods could provide incremental values. Continuing this way yields consistent values which are approximative to \( 1.0048 \) itself after confirming limited impact on higher order terms.
Final answer:
\[
\boxed{1.0048}
\]
We want to calculate \( (2 + \sqrt{3})^6 + (2 - \sqrt{3})^6 \).
Let \( a = 2 + \sqrt{3} \) and \( b = 2 - \sqrt{3} \). Notice that \( a \) and \( b \) are conjugates, and their product is:
\[
ab = (2 + \sqrt{3})(2 - \sqrt{3}) = 4 - 3 = 1
\]
This indicates that \( b = \frac{1}{a} \). Therefore, we can express \( b^6 = \left(\frac{1}{a}\right)^6 = \frac{1}{a^6} \).
Now, we want to find:
\[
a^6 + b^6 = a^6 + \frac{1}{a^6}
\]
Next, we can express \( a^6 + \frac{1}{a^6} \) using the identity:
\[
x^6 + \frac{1}{x^6} = \left(x^3 + \frac{1}{x^3}\right)^2 - 2
\]
First, we need to find \( a + b \) and \( ab \).
\[
a + b = (2 + \sqrt{3}) + (2 - \sqrt{3}) = 4
\]
\[
ab = 1
\]
Next, we can find \( a^n + b^n \) using the recurrence relation:
\[
s_n = a^n + b^n
\]
with initial conditions \( s_0 = 2 \), \( s_1 = 4 \) and the recurrence relation:
\[
s_n = (a + b)s_{n-1} - ab s_{n-2} = 4s_{n-1} - s_{n-2}
\]
Now we calculate \( s_n \):
- \( s_2 = 4s_1 - s_0 = 4 \cdot 4 - 2 = 16 - 2 = 14 \)
- \( s_3 = 4s_2 - s_1 = 4 \cdot 14 - 4 = 56 - 4 = 52 \)
- \( s_4 = 4s_3 - s_2 = 4 \cdot 52 - 14 = 208 - 14 = 194 \)
- \( s_5 = 4s_4 - s_3 = 4 \cdot 194 - 52 = 776 - 52 = 724 \)
- \( s_6 = 4s_5 - s_4 = 4 \cdot 724 - 194 = 2896 - 194 = 2702 \)
Thus, the value of \( (2+\sqrt{3})^6 + (2-\sqrt{3})^6 \) is:
\[
\boxed{2702}
\]
---
### Problem 2:
We need to evaluate \( 1.0048 \) using the binomial theorem, writing it as \( 1 + 0.0048 \), and apply the binomial expansion for \( (1 + x)^n \).
Using the binomial expansion,
\[
(1 + x)^n = \sum_{k=0}^{n} \binom{n}{k} x^k
\]
We choose \( n = 1 \) and \( x = 0.0048 \):
\[
(1.0048)^{1} = 1 + \binom{1}{1} (0.0048) = 1 + 0.0048 = 1.0048
\]
However, let's consider \( n = \frac{1}{2} \) for demonstration. We'll try evaluating it with \( n = 1 + \frac{1}{4} \) to find an approximation:
To expand to five decimal places and for greater accuracy, using up to the \( x^4 \) term:
\[
(1 + x)^{n} \approx 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \frac{n(n-1)(n-2)(n-3)}{4!}x^4
\]
Let \( n = 1 \) and \( x = 0.0048 \):
1. \( 1 \)
2. \( nx = 1 \times 0.0048 = 0.0048 \)
3. \( \frac{1 \cdot 0}{2} \cdot (0.0048)^2 = 0 \)
4. \( \frac{1 \cdot 0 \cdot -1}{3} \cdot (0.0048)^3 = 0 \)
5. \( \frac{1 \cdot 0 \cdot -1 \cdot -2}{4!} \cdot (0.0048)^4 = 0 \)
Taking the above terms into account:
Thus the result is:
\[
1.0048
\]
For more precision, stop here; otherwise, numerical methods could provide incremental values. Continuing this way yields consistent values which are approximative to \( 1.0048 \) itself after confirming limited impact on higher order terms.
Final answer:
\[
\boxed{1.0048}
\]