Question
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
To solve the first problem \((2 + \sqrt{3})^6 + (2 - \sqrt{3})^6\), we can use the properties of the binomial theorem and symmetry.
Using the binomial expansion, we have:
\[
(2 + \sqrt{3})^6 = \sum_{k=0}^{6} \binom{6}{k} (2)^{6-k} (\sqrt{3})^k
\]
\[
(2 - \sqrt{3})^6 = \sum_{k=0}^{6} \binom{6}{k} (2)^{6-k} (-\sqrt{3})^k
\]
When we add these two expansions together:
\[
(2 + \sqrt{3})^6 + (2 - \sqrt{3})^6 = \sum_{k=0}^{6} \binom{6}{k} (2)^{6-k} ((\sqrt{3})^k + (-\sqrt{3})^k)
\]
Observe that for even \(k\), \((\sqrt{3})^k + (-\sqrt{3})^k\) is \(2(\sqrt{3})^k\) and for odd \(k\), it is 0. Thus, we only need to consider the even terms in the series:
- For \(k = 0\): \(\binom{6}{0} (2)^6 = 64\)
- For \(k = 2\): \(\binom{6}{2} (2)^4 (\sqrt{3})^2 = 15 \cdot 16 \cdot 3 = 720\)
- For \(k = 4\): \(\binom{6}{4} (2)^2 (\sqrt{3})^4 = 15 \cdot 4 \cdot 9 = 540\)
- For \(k = 6\): \(\binom{6}{6} (2)^0 (\sqrt{3})^6 = 1 \cdot 27 = 27\)
Now we add these contributions together:
\[
(2 + \sqrt{3})^6 + (2 - \sqrt{3})^6 = 64 + 720 + 540 + 27
\]
\[
= 64 + 720 + 540 + 27 = 1351
\]
Thus, the final value is:
\[
\boxed{1351}
\]
---
For the second part, we want to evaluate \(1.0048\) using the binomial theorem, and we can express it as:
\[
1.0048 = 1 + 0.0048
\]
Using the binomial series expansion for \((1 + x)^n\):
\[
(1 + x)^n = 1 + nx + \frac{n(n-1)x^2}{2!} + \frac{n(n-1)(n-2)x^3}{3!} + \ldots
\]
For small values of \(x\), we can approximate:
Let \(x = 0.0048\) and \(n = 1\):
\[
(1 + 0.0048)^1 \approx 1 + 1(0.0048) = 1 + 0.0048 = 1.0048
\]
To get a finer approximation, we could include the square term, which is appropriate since we want a more precise value. If we, for instance, consider the limit of accuracy we want (5 decimal places) and let’s continue on adding it with \(n=1\):
\[
1 + 1(0.0048) + \frac{1 \cdot 0.0048^2}{2} \approx 1.0048 + \frac{0.00002304}{2} \approx 1.0048 + 0.00001152
\]
Calculating this gives:
\[
1.0048 + 0.00001152 = 1.00481152
\]
Thus, rounding this to 5 decimal places gives:
\[
\boxed{1.00481}
\]
Using the binomial expansion, we have:
\[
(2 + \sqrt{3})^6 = \sum_{k=0}^{6} \binom{6}{k} (2)^{6-k} (\sqrt{3})^k
\]
\[
(2 - \sqrt{3})^6 = \sum_{k=0}^{6} \binom{6}{k} (2)^{6-k} (-\sqrt{3})^k
\]
When we add these two expansions together:
\[
(2 + \sqrt{3})^6 + (2 - \sqrt{3})^6 = \sum_{k=0}^{6} \binom{6}{k} (2)^{6-k} ((\sqrt{3})^k + (-\sqrt{3})^k)
\]
Observe that for even \(k\), \((\sqrt{3})^k + (-\sqrt{3})^k\) is \(2(\sqrt{3})^k\) and for odd \(k\), it is 0. Thus, we only need to consider the even terms in the series:
- For \(k = 0\): \(\binom{6}{0} (2)^6 = 64\)
- For \(k = 2\): \(\binom{6}{2} (2)^4 (\sqrt{3})^2 = 15 \cdot 16 \cdot 3 = 720\)
- For \(k = 4\): \(\binom{6}{4} (2)^2 (\sqrt{3})^4 = 15 \cdot 4 \cdot 9 = 540\)
- For \(k = 6\): \(\binom{6}{6} (2)^0 (\sqrt{3})^6 = 1 \cdot 27 = 27\)
Now we add these contributions together:
\[
(2 + \sqrt{3})^6 + (2 - \sqrt{3})^6 = 64 + 720 + 540 + 27
\]
\[
= 64 + 720 + 540 + 27 = 1351
\]
Thus, the final value is:
\[
\boxed{1351}
\]
---
For the second part, we want to evaluate \(1.0048\) using the binomial theorem, and we can express it as:
\[
1.0048 = 1 + 0.0048
\]
Using the binomial series expansion for \((1 + x)^n\):
\[
(1 + x)^n = 1 + nx + \frac{n(n-1)x^2}{2!} + \frac{n(n-1)(n-2)x^3}{3!} + \ldots
\]
For small values of \(x\), we can approximate:
Let \(x = 0.0048\) and \(n = 1\):
\[
(1 + 0.0048)^1 \approx 1 + 1(0.0048) = 1 + 0.0048 = 1.0048
\]
To get a finer approximation, we could include the square term, which is appropriate since we want a more precise value. If we, for instance, consider the limit of accuracy we want (5 decimal places) and let’s continue on adding it with \(n=1\):
\[
1 + 1(0.0048) + \frac{1 \cdot 0.0048^2}{2} \approx 1.0048 + \frac{0.00002304}{2} \approx 1.0048 + 0.00001152
\]
Calculating this gives:
\[
1.0048 + 0.00001152 = 1.00481152
\]
Thus, rounding this to 5 decimal places gives:
\[
\boxed{1.00481}
\]