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Question Number 188417 by 073 last updated on 01/Mar/23

Commented by SEKRET last updated on 01/Mar/23

D

$$\:\boldsymbol{\mathrm{D}} \\ $$

Answered by SEKRET last updated on 01/Mar/23

Σ_(k=1) ^(10) (((k−1)!∙k∙(k+1)....(k+10))/((k−1)!))=? Σ_(k=1) ^(10) (((k+10)!)/((k−1)!))=Σ_(k=1) ^(10) (((k+10)!∙(k+11−(k−1)))/(12∙(k−1)!)) Σ_(k=1) ^(10) (((k+11)! −(k+10)!∙(k−1))/(12∙(k−1)!)) = =(1/(12))∙Σ_(k=1) ^(10) ((((k+11)!)/((k−1)!)) − (((k+10)!∙(k−1))/((k−1)!)))= = (1/(12))∙(((12!)/(0!))−((11!∙0)/(0!))+((13!)/(1!))−((12!)/(0!))+((14!)/(2!))−((13!)/(1!))+...+((21!)/(9!))−((20!)/(8!))) =(1/(12))∙(((12!)/(0!))+((13!)/(1!))−((12!)/(0!))+((14!)/(2!))−((13!)/(1!))+...+((21!)/(9!))−((20!)/(8!))) = ((21!)/(12∙9!)) ABDULAZIZ ABDUVALIYEV

$$\:\:\underset{\boldsymbol{\mathrm{k}}=\mathrm{1}} {\overset{\mathrm{10}} {\sum}}\frac{\left(\boldsymbol{\mathrm{k}}−\mathrm{1}\right)!\centerdot\boldsymbol{\mathrm{k}}\centerdot\left(\boldsymbol{\mathrm{k}}+\mathrm{1}\right)....\left(\boldsymbol{\mathrm{k}}+\mathrm{10}\right)}{\left(\boldsymbol{\mathrm{k}}−\mathrm{1}\right)!}=? \\ $$$$\:\:\underset{\boldsymbol{\mathrm{k}}=\mathrm{1}} {\overset{\mathrm{10}} {\sum}}\:\frac{\left(\boldsymbol{\mathrm{k}}+\mathrm{10}\right)!}{\left(\boldsymbol{\mathrm{k}}−\mathrm{1}\right)!}=\underset{\boldsymbol{\mathrm{k}}=\mathrm{1}} {\overset{\mathrm{10}} {\sum}}\frac{\left(\boldsymbol{\mathrm{k}}+\mathrm{10}\right)!\centerdot\left(\boldsymbol{\mathrm{k}}+\mathrm{11}−\left(\boldsymbol{\mathrm{k}}−\mathrm{1}\right)\right)}{\mathrm{12}\centerdot\left(\boldsymbol{\mathrm{k}}−\mathrm{1}\right)!} \\ $$$$\:\underset{\boldsymbol{\mathrm{k}}=\mathrm{1}} {\overset{\mathrm{10}} {\sum}}\:\frac{\left(\boldsymbol{\mathrm{k}}+\mathrm{11}\right)!\:−\left(\boldsymbol{\mathrm{k}}+\mathrm{10}\right)!\centerdot\left(\boldsymbol{\mathrm{k}}−\mathrm{1}\right)}{\mathrm{12}\centerdot\left(\boldsymbol{\mathrm{k}}−\mathrm{1}\right)!}\:= \\ $$$$=\frac{\mathrm{1}}{\mathrm{12}}\centerdot\underset{\boldsymbol{\mathrm{k}}=\mathrm{1}} {\overset{\mathrm{10}} {\sum}}\:\left(\frac{\left(\boldsymbol{\mathrm{k}}+\mathrm{11}\right)!}{\left(\boldsymbol{\mathrm{k}}−\mathrm{1}\right)!}\:−\:\frac{\left(\boldsymbol{\mathrm{k}}+\mathrm{10}\right)!\centerdot\left(\boldsymbol{\mathrm{k}}−\mathrm{1}\right)}{\left(\boldsymbol{\mathrm{k}}−\mathrm{1}\right)!}\right)= \\ $$$$\:=\:\frac{\mathrm{1}}{\mathrm{12}}\centerdot\left(\frac{\mathrm{12}!}{\mathrm{0}!}−\frac{\mathrm{11}!\centerdot\mathrm{0}}{\mathrm{0}!}+\frac{\mathrm{13}!}{\mathrm{1}!}−\frac{\mathrm{12}!}{\mathrm{0}!}+\frac{\mathrm{14}!}{\mathrm{2}!}−\frac{\mathrm{13}!}{\mathrm{1}!}+...+\frac{\mathrm{21}!}{\mathrm{9}!}−\frac{\mathrm{20}!}{\mathrm{8}!}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{12}}\centerdot\left(\frac{\mathrm{12}!}{\mathrm{0}!}+\frac{\mathrm{13}!}{\mathrm{1}!}−\frac{\mathrm{12}!}{\mathrm{0}!}+\frac{\mathrm{14}!}{\mathrm{2}!}−\frac{\mathrm{13}!}{\mathrm{1}!}+...+\frac{\mathrm{21}!}{\mathrm{9}!}−\frac{\mathrm{20}!}{\mathrm{8}!}\right) \\ $$$$=\:\frac{\mathrm{21}!}{\mathrm{12}\centerdot\mathrm{9}!} \\ $$$$\:\:\boldsymbol{{ABDULAZIZ}}\:\:\:\boldsymbol{{ABDUVALIYEV}} \\ $$$$ \\ $$

Commented by 073 last updated on 02/Mar/23

nice solution

$$\mathrm{nice}\:\mathrm{solution} \\ $$

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