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Question Number 178500 by infinityaction last updated on 17/Oct/22

Commented by abdullahoudou last updated on 17/Oct/22

it would be nice to give the rest of questions instead of only number 15 ...

$${it}\:{would}\:{be}\:{nice}\:{to}\:{give}\:{the}\:{rest}\:{of}\:{questions}\:{instead}\:{of}\:{only}\:{number}\:\mathrm{15}\:... \\ $$

Answered by mindispower last updated on 17/Oct/22

==Σ_(k≥0) ((k(k−1)n!)/((n−k)!k!))  =Σ_(k=0) ^n k(k−1)C_k ^n   Σ_(k=0) ^n x^k C_n ^k =(1+x)^n   ⇒Σ_(k≥1) kx^(k−1) C_k ^n =n(1+x)^(n−1)   Σk(k−1)x^(k−2) =n(n−1)(1+x)^(n−2)   Σk(k−1)C_n ^k =Σ((n!)/((n−k)!(k−2)!))=n(n−1)2^(n−2)

$$==\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{{k}\left({k}−\mathrm{1}\right){n}!}{\left({n}−{k}\right)!{k}!} \\ $$$$=\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{k}\left({k}−\mathrm{1}\right){C}_{{k}} ^{{n}} \\ $$$$\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{x}^{{k}} {C}_{{n}} ^{{k}} =\left(\mathrm{1}+{x}\right)^{{n}} \\ $$$$\Rightarrow\underset{{k}\geqslant\mathrm{1}} {\sum}{kx}^{{k}−\mathrm{1}} {C}_{{k}} ^{{n}} ={n}\left(\mathrm{1}+{x}\right)^{{n}−\mathrm{1}} \\ $$$$\Sigma{k}\left({k}−\mathrm{1}\right){x}^{{k}−\mathrm{2}} ={n}\left({n}−\mathrm{1}\right)\left(\mathrm{1}+{x}\right)^{{n}−\mathrm{2}} \\ $$$$\Sigma{k}\left({k}−\mathrm{1}\right){C}_{{n}} ^{{k}} =\Sigma\frac{{n}!}{\left({n}−{k}\right)!\left({k}−\mathrm{2}\right)!}={n}\left({n}−\mathrm{1}\right)\mathrm{2}^{{n}−\mathrm{2}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by infinityaction last updated on 17/Oct/22

thank you sir

$${thank}\:{you}\:{sir} \\ $$

Commented by mindispower last updated on 02/Nov/22

withe pleasur

$${withe}\:{pleasur} \\ $$

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