The sum of all rational terms in the expansion of $(2+\sqrt{3})^8$ is
18817
To find the sum of all rational terms in the expansion of $(2+\sqrt{3})^8$, we need to identify the terms that are purely numerical, without any irrational components like $\sqrt{3}$.
The general term in the binomial expansion of $(a+b)^n$ is given by the formula:
$ T_{k+1} = \binom{n}{k} a^{n-k} b^k $
For the given expression $(2+\sqrt{3})^8$, we have $a=2$, $b=\sqrt{3}$, and $n=8$. Substituting these values, the general term becomes:
$ T_{k+1} = \binom{8}{k} 2^{8-k} (\sqrt{3})^k $
A term $T_{k+1}$ is rational if the part involving $\sqrt{3}$ is rational. This occurs when the exponent $k$ is an even number, because $(\sqrt{3})^k = (3^{1/2})^k = 3^{k/2}$, which is rational only if $k/2$ is an integer (i.e., $k$ is even).
In the expansion of $(2+\sqrt{3})^8$, the possible values for $k$ range from 0 to 8. The rational terms occur for the following even values of $k$: $0, 2, 4, 6, 8$.
We calculate the value of each rational term:
| Index $k$ | Term Calculation ($ \binom{8}{k} 2^{8-k} (\sqrt{3})^k $) | Term Value |
| $k=0$ | $ \binom{8}{0} 2^{8-0} (\sqrt{3})^0 = 1 \times 2^8 \times 1 $ | $ 256 $ |
| $k=2$ | $ \binom{8}{2} 2^{8-2} (\sqrt{3})^2 = 28 \times 2^6 \times 3^1 $ | $ 28 \times 64 \times 3 = 5376 $ |
| $k=4$ | $ \binom{8}{4} 2^{8-4} (\sqrt{3})^4 = 70 \times 2^4 \times 3^2 $ | $ 70 \times 16 \times 9 = 10080 $ |
| $k=6$ | $ \binom{8}{6} 2^{8-6} (\sqrt{3})^6 = 28 \times 2^2 \times 3^3 $ | $ 28 \times 4 \times 27 = 3024 $ |
| $k=8$ | $ \binom{8}{8} 2^{8-8} (\sqrt{3})^8 = 1 \times 2^0 \times 3^4 $ | $ 1 \times 1 \times 81 = 81 $ |
The sum $S$ of all rational terms is obtained by adding the values calculated above:
$ S = 256 + 5376 + 10080 + 3024 + 81 $
Adding these values gives:
$ S = 18817 $
Thus, the sum of all rational terms in the expansion of $(2+\sqrt{3})^8$ is 18817.
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Let the length of the conjugate axis of H be $p$ and the length of its latus rectum be $q$. Then $p^2 + q^2$ is equal to
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