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)

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

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

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

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

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

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

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

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

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1.

$T_{k-1}<T_k$

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

$frac{1}{(k-1)!cdot(21-k)!}cdotsqrt2< frac{1}{k!cdot(20-k)!}sqrt5$

$frac{1}{(k-1)!cdot(20-k)! cdot (21-k)}cdotsqrt2< frac{1}{(k-1)! cdot kcdot(20-k)!}sqrt5 /cdot((k-1)!cdot(20-k)! )$

(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.

$T_k>T_{k+1}$

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

$frac{1}{k!cdot(20-k)!}cdotsqrt2>frac{1}{(k+1)!cdot(19-k)!} sqrt5$

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

(1<k<20)

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

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

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

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

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

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

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

1,2

$k=12$

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

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

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