Question

Как найти наибольший член разложения бинома?
$(\sqrt{5} +\sqrt{2} )^{20}$

Answer 1

$(\sqrt{5} +\sqrt{2} )^{20}$

$T_k$ - наибольший член (1<k<20)

$T_k= {20 \choose k} (\sqrt5)^{k}\cdot(\sqrt2)^{20-k}$

$T_k= \frac{20!}{k!\cdot(20-k)!}(\sqrt5)^{k}\cdot(\sqrt2)^{20-k}$

---------------

$T_{k-1}= {20 \choose k-1} (\sqrt5)^{k-1}\cdot(\sqrt2)^{20-k+1}$

$T_{k-1}=\frac{20!}{(k-1)!\cdot(20-k+1)!}(\sqrt5)^{k-1}\cdot(\sqrt2)^{21-k}$

$T_{k-1}=\frac{20!}{(k-1)!\cdot(21-k)!}(\sqrt5)^{k-1}\cdot(\sqrt2)^{21-k}$

---------------

$T_{k+1}= {20 \choose k+1} (\sqrt5)^{k+1}\cdot(\sqrt2)^{20-k-1}$

$T_{k+1}=\frac{20!}{(k+1)!\cdot(20-k-1)!} (\sqrt5)^{k+1}\cdot(\sqrt2)^{19-k}$

$T_{k+1}=\frac{20!}{(k+1)!\cdot(19-k)!} (\sqrt5)^{k+1}\cdot(\sqrt2)^{19-k}$

---------------

1.

$T_{k-1}<T_k$

$\frac{20!}{(k-1)!\cdot(21-k)!}(\sqrt5)^{k-1}\cdot(\sqrt2)^{21-k}< \frac{20!}{k!\cdot(20-k)!}(\sqrt5)^{k}\cdot(\sqrt2)^{20-k}\ /:(20! \cdot ( \sqrt{5} )^{k-1} \cdot ( \sqrt{2} )^{20-k})$

$\frac{1}{(k-1)!\cdot(21-k)!}\cdot\sqrt2< \frac{1}{k!\cdot(20-k)!}\sqrt5$

$\frac{1}{(k-1)!\cdot(20-k)! \cdot (21-k)}\cdot\sqrt2< \frac{1}{(k-1)! \cdot k\cdot(20-k)!}\sqrt5\ /\cdot((k-1)!\cdot(20-k)! )$

(1<k<20)

$\frac{\sqrt2}{21-k}< \frac{\sqrt5}{k}\ /\cdotk(21-k)$

$\sqrt2k< \sqrt{5}(21-k)$

$\sqrt2k<21 \sqrt{5}- \sqrt{5} k$

$\sqrt2k+\sqrt{5} k<21 \sqrt{5}$

$(\sqrt2+\sqrt{5}) k<21 \sqrt{5}\ /:(\sqrt2+\sqrt{5})$

$k< \frac{21 \sqrt{5}}{\sqrt2+\sqrt{5}}$

$\frac{21 \sqrt{5}}{\sqrt2+\sqrt{5}} \approx 12,86$

2.

$T_k>T_{k+1}$

$\frac{20!}{k!\cdot(20-k)!}(\sqrt5)^{k}\cdot(\sqrt2)^{20-k}>\frac{20!}{(k+1)!\cdot(19-k)!} (\sqrt5)^{k+1}\cdot(\sqrt2)^{19-k}\ /:(20! \cdot ( \sqrt{5} )^k \cdot ( \sqrt{2} )^{19-k})$

$\frac{1}{k!\cdot(20-k)!}\cdot\sqrt2>\frac{1}{(k+1)!\cdot(19-k)!} \sqrt5$

$\frac{\sqrt2}{k!\cdot(19-k)!(20-k)}>\frac{\sqrt5}{k!(k+1)\cdot(19-k)!}\ /\cdot k!\cdot(19-k)!$

(1<k<20)

$\frac{\sqrt2}{20-k}>\frac{\sqrt5}{k+1}\ /\cdot(20-k)k$

$\sqrt2(k+1)>\sqrt5(20-k)$

$\sqrt2k+ \sqrt{2}>20\sqrt5- \sqrt{5} k$

$\sqrt2k+\sqrt{5} k>20\sqrt5-\sqrt{2}$

$(\sqrt2+\sqrt{5}) k>20\sqrt5-\sqrt{2$

$k> \frac{20\sqrt5-\sqrt{2}}{\sqrt2+\sqrt{5}}$

$\frac{20\sqrt5-\sqrt{2}}{\sqrt2+\sqrt{5}} \approx11,86$

1,2

$k=12$

$T_{12}= \frac{20!}{12!\cdot(20-12)!}(\sqrt5)^{12}\cdot(\sqrt2)^{20-12}$

$T_{12}= \frac{20!}{12!\cdot8!} \cdot 5^6\cdot(\sqrt2)^{8}$

$T_{12}= \frac{20!}{12!\cdot8!} \cdot 5^6\cdot2^4$