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Question Number 148333 by mathdanisur last updated on 27/Jul/21

((x)^(1/3) + (1/( (√x))))^(15) Find the limit that does not inclued the variable x in the opening of the binomial.

$$\left(\sqrt[{\mathrm{3}}]{{x}}\:+\:\frac{\mathrm{1}}{\:\sqrt{{x}}}\right)^{\mathrm{15}} \\ $$$${Find}\:{the}\:{limit}\:{that}\:{does}\:{not}\:{inclued} \\ $$$${the}\:{variable}\:\boldsymbol{{x}}\:{in}\:{the}\:{opening}\:{of}\:{the} \\ $$$${binomial}. \\ $$

Answered by qaz last updated on 27/Jul/21

(((15)),(( 6)) )

$$\begin{pmatrix}{\mathrm{15}}\\{\:\mathrm{6}}\end{pmatrix} \\ $$

Commented by mathdanisur last updated on 27/Jul/21

How Sir

$${How}\:{Sir} \\ $$

Answered by mindispower last updated on 27/Jul/21

(a+b)^n =Σ_(k=0) ^n C_n ^k a^k b^(n−k) a=x^(1/3) ,x^(−(1/2)) ,n=15 =Σ_(k=0) ^n C_n ^k x^(k/3) x^(−(1/2)(15−k)) ⇒(k/3)−(1/2)(15−k)=0 ⇒2k−45+3k=0 k=9 the constante is C_9 ^(15) =C_(15−9) ^(15) =C_6 ^(15)

$$\left({a}+{b}\right)^{{n}} =\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{C}_{{n}} ^{{k}} {a}^{{k}} {b}^{{n}−{k}} \\ $$$${a}={x}^{\frac{\mathrm{1}}{\mathrm{3}}} ,{x}^{−\frac{\mathrm{1}}{\mathrm{2}}} ,{n}=\mathrm{15} \\ $$$$=\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{C}_{{n}} ^{{k}} {x}^{\frac{{k}}{\mathrm{3}}} {x}^{−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{15}−{k}\right)} \\ $$$$\Rightarrow\frac{{k}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{15}−{k}\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{2}{k}−\mathrm{45}+\mathrm{3}{k}=\mathrm{0} \\ $$$${k}=\mathrm{9} \\ $$$${the}\:{constante}\:{is}\:{C}_{\mathrm{9}} ^{\mathrm{15}} ={C}_{\mathrm{15}−\mathrm{9}} ^{\mathrm{15}} ={C}_{\mathrm{6}} ^{\mathrm{15}} \\ $$

Commented by mathdanisur last updated on 27/Jul/21

Thank you Ser, answer: 9.?

$${Thank}\:{you}\:{Ser},\:{answer}:\:\mathrm{9}.? \\ $$

Commented by qaz last updated on 27/Jul/21

(x)^(1/3) +(1/( (√x)))=x^(1/3) +x^(−(1/2)) { (((1/3)a−(1/2)b=0)),((a+b=15)) :} ⇒ { ((a=9)),((b=6)) :} So canstant coefficient is (((15)),(( 6)) ).

$$\sqrt[{\mathrm{3}}]{\mathrm{x}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{x}}}=\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{3}}} +\mathrm{x}^{−\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\begin{cases}{\frac{\mathrm{1}}{\mathrm{3}}\mathrm{a}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{b}=\mathrm{0}}\\{\mathrm{a}+\mathrm{b}=\mathrm{15}}\end{cases} \\ $$$$\Rightarrow\begin{cases}{\mathrm{a}=\mathrm{9}}\\{\mathrm{b}=\mathrm{6}}\end{cases} \\ $$$$\mathrm{So}\:\mathrm{canstant}\:\mathrm{coefficient}\:\mathrm{is}\:\begin{pmatrix}{\mathrm{15}}\\{\:\mathrm{6}}\end{pmatrix}. \\ $$

Commented by mathdanisur last updated on 27/Jul/21

Thank you Sir

$${Thank}\:{you}\:{Sir} \\ $$

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