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C-0-2C-1-3C-2-n-1-C-n-n-2-2-n-1-using-Bionomial-teorm-




Question Number 25969 by soyebshaikh41@gmail.com last updated on 17/Dec/17
C_0 +2C_1 +3C_2 +..........+(n+1)C_n =(n+2)2^(n−1)   using  Bionomial teorm
$$\mathrm{C}_{\mathrm{0}} +\mathrm{2C}_{\mathrm{1}} +\mathrm{3C}_{\mathrm{2}} +……….+\left(\mathrm{n}+\mathrm{1}\right)\mathrm{C}_{\mathrm{n}} =\left(\mathrm{n}+\mathrm{2}\right)\mathrm{2}^{\mathrm{n}−\mathrm{1}} \\ $$$$\mathrm{using}\:\:\mathrm{Bionomial}\:\mathrm{teorm} \\ $$
Commented by moxhix last updated on 17/Dec/17
show _n C_0 +2 _n C_1 +3 _n C_2 +...+(n+1) _n C_n =(n+2)2^(n−1)   put f(x)=x(1+x)^n     f ′(x)=(1+x)^n +nx(1+x)^(n−1) =(n+1+x)(1+x)^(n−1)   f ′(1)=(n+2)2^(n−1)     f(x)=xΣ_(k=0) ^n  _n C_k x^k =Σ_(k=0) ^n  _n C_k x^(k+1)   f ′(x)=Σ_(k=0) ^n  _n C_k (k+1)x^k   f ′(1)=Σ_(k=0) ^n  _n C_k (k+1)    ∴ _n C_0 +2 _n C_1 +3 _n C_2 +...+(n+1) _n C_n =(n+2)2^(n−1)
$${show}\:_{{n}} {C}_{\mathrm{0}} +\mathrm{2}\:_{{n}} {C}_{\mathrm{1}} +\mathrm{3}\:_{{n}} {C}_{\mathrm{2}} +…+\left({n}+\mathrm{1}\right)\:_{{n}} {C}_{{n}} =\left({n}+\mathrm{2}\right)\mathrm{2}^{{n}−\mathrm{1}} \\ $$$${put}\:{f}\left({x}\right)={x}\left(\mathrm{1}+{x}\right)^{{n}} \\ $$$$ \\ $$$${f}\:'\left({x}\right)=\left(\mathrm{1}+{x}\right)^{{n}} +{nx}\left(\mathrm{1}+{x}\right)^{{n}−\mathrm{1}} =\left({n}+\mathrm{1}+{x}\right)\left(\mathrm{1}+{x}\right)^{{n}−\mathrm{1}} \\ $$$${f}\:'\left(\mathrm{1}\right)=\left({n}+\mathrm{2}\right)\mathrm{2}^{{n}−\mathrm{1}} \\ $$$$ \\ $$$${f}\left({x}\right)={x}\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\:_{{n}} {C}_{{k}} {x}^{{k}} =\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\:_{{n}} {C}_{{k}} {x}^{{k}+\mathrm{1}} \\ $$$${f}\:'\left({x}\right)=\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\:_{{n}} {C}_{{k}} \left({k}+\mathrm{1}\right){x}^{{k}} \\ $$$${f}\:'\left(\mathrm{1}\right)=\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\:_{{n}} {C}_{{k}} \left({k}+\mathrm{1}\right) \\ $$$$ \\ $$$$\therefore\:_{{n}} {C}_{\mathrm{0}} +\mathrm{2}\:_{{n}} {C}_{\mathrm{1}} +\mathrm{3}\:_{{n}} {C}_{\mathrm{2}} +…+\left({n}+\mathrm{1}\right)\:_{{n}} {C}_{{n}} =\left({n}+\mathrm{2}\right)\mathrm{2}^{{n}−\mathrm{1}} \\ $$

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