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Question Number 151073 by mathdanisur last updated on 18/Aug/21

the value of   (((50)),(0) )^2 + (((50)),(1) )^2 + (((50)),(2) )^2 +...+ (((50)),((49)) )^2 + (((50)),((50)) )^2

$$\mathrm{the}\:\mathrm{value}\:\mathrm{of} \\ $$$$\begin{pmatrix}{\mathrm{50}}\\{\mathrm{0}}\end{pmatrix}^{\mathrm{2}} +\begin{pmatrix}{\mathrm{50}}\\{\mathrm{1}}\end{pmatrix}^{\mathrm{2}} +\begin{pmatrix}{\mathrm{50}}\\{\mathrm{2}}\end{pmatrix}^{\mathrm{2}} +...+\begin{pmatrix}{\mathrm{50}}\\{\mathrm{49}}\end{pmatrix}^{\mathrm{2}} +\begin{pmatrix}{\mathrm{50}}\\{\mathrm{50}}\end{pmatrix}^{\mathrm{2}} \\ $$

Answered by ArielVyny last updated on 18/Aug/21

S_n =Σ_(k=0) ^(50) (C_(50) ^k )^2   ln(S_n )=2Σ_(k=0) ^(50) ln(((50!)/(k!(50−k)!)))=2ln(50!)(50+1)−2Σ_(k=0) ^(50) ln[k!(50−k)!]  ln(S_n )=102ln(50!)−2Σ_(k=0) ^(50) ln(k!)−2Σ_(k=0) ^(50) ln(50−k)!    ln(S_n )=102ln(50!)→S_n =e^(102ln(50!)) =e^(ln(50!)^(102) )   S_n =(50!)^(102)

$${S}_{{n}} =\underset{{k}=\mathrm{0}} {\overset{\mathrm{50}} {\sum}}\left({C}_{\mathrm{50}} ^{{k}} \right)^{\mathrm{2}} \\ $$$${ln}\left({S}_{{n}} \right)=\mathrm{2}\underset{{k}=\mathrm{0}} {\overset{\mathrm{50}} {\sum}}{ln}\left(\frac{\mathrm{50}!}{{k}!\left(\mathrm{50}−{k}\right)!}\right)=\mathrm{2}{ln}\left(\mathrm{50}!\right)\left(\mathrm{50}+\mathrm{1}\right)−\mathrm{2}\underset{{k}=\mathrm{0}} {\overset{\mathrm{50}} {\sum}}{ln}\left[{k}!\left(\mathrm{50}−{k}\right)!\right] \\ $$$${ln}\left({S}_{{n}} \right)=\mathrm{102}{ln}\left(\mathrm{50}!\right)−\mathrm{2}\underset{{k}=\mathrm{0}} {\overset{\mathrm{50}} {\sum}}{ln}\left({k}!\right)−\mathrm{2}\underset{{k}=\mathrm{0}} {\overset{\mathrm{50}} {\sum}}{ln}\left(\mathrm{50}−{k}\right)! \\ $$$$ \\ $$$${ln}\left({S}_{{n}} \right)=\mathrm{102}{ln}\left(\mathrm{50}!\right)\rightarrow{S}_{{n}} ={e}^{\mathrm{102}{ln}\left(\mathrm{50}!\right)} ={e}^{{ln}\left(\mathrm{50}!\right)^{\mathrm{102}} } \\ $$$${S}_{{n}} =\left(\mathrm{50}!\right)^{\mathrm{102}} \\ $$

Commented by mathdanisur last updated on 18/Aug/21

Thank you Ser, answer:  (((100)),((50)) )

$$\mathrm{Thank}\:\mathrm{you}\:\boldsymbol{\mathrm{S}}\mathrm{er},\:\mathrm{answer}:\:\begin{pmatrix}{\mathrm{100}}\\{\mathrm{50}}\end{pmatrix} \\ $$

Commented by mr W last updated on 18/Aug/21

wrong!  S=a+b  ln S=ln (a+b)≠ln a+ln b

$${wrong}! \\ $$$${S}={a}+{b} \\ $$$$\mathrm{ln}\:{S}=\mathrm{ln}\:\left({a}+{b}\right)\neq\mathrm{ln}\:{a}+\mathrm{ln}\:{b} \\ $$

Answered by mr W last updated on 18/Aug/21

in (1+x)^n (1+x)^n   the coef. of x^n  is Σ_(k=0) ^n C_k ^n C_(n−k) ^n =Σ_(k=0) ^n (C_k ^n )^2   but (1+x)^n (1+x)^n =(1+x)^(2n)  and the  the coef. of x^n  is C_n ^(2n) . therefore  Σ_(k=0) ^n (C_k ^n )^2 =C_n ^(2n)   ⇒(C_0 ^(50) )^2 +(C_1 ^(50) )^2 +...+(C_(50) ^(50) )^2 =C_(50) ^(100)

$${in}\:\left(\mathrm{1}+{x}\right)^{{n}} \left(\mathrm{1}+{x}\right)^{{n}} \\ $$$${the}\:{coef}.\:{of}\:{x}^{{n}} \:{is}\:\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{C}_{{k}} ^{{n}} {C}_{{n}−{k}} ^{{n}} =\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\left({C}_{{k}} ^{{n}} \right)^{\mathrm{2}} \\ $$$${but}\:\left(\mathrm{1}+{x}\right)^{{n}} \left(\mathrm{1}+{x}\right)^{{n}} =\left(\mathrm{1}+{x}\right)^{\mathrm{2}{n}} \:{and}\:{the} \\ $$$${the}\:{coef}.\:{of}\:{x}^{{n}} \:{is}\:{C}_{{n}} ^{\mathrm{2}{n}} .\:{therefore} \\ $$$$\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\left({C}_{{k}} ^{{n}} \right)^{\mathrm{2}} ={C}_{{n}} ^{\mathrm{2}{n}} \\ $$$$\Rightarrow\left({C}_{\mathrm{0}} ^{\mathrm{50}} \right)^{\mathrm{2}} +\left({C}_{\mathrm{1}} ^{\mathrm{50}} \right)^{\mathrm{2}} +...+\left({C}_{\mathrm{50}} ^{\mathrm{50}} \right)^{\mathrm{2}} ={C}_{\mathrm{50}} ^{\mathrm{100}} \\ $$

Commented by mathdanisur last updated on 18/Aug/21

Thank you Ser

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{Ser} \\ $$

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