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

Commented by SEKRET last updated on 01/Mar/23

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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

∑10k=1(k−1)!⋅k⋅(k+1)....(k+10)(k−1)!=?∑10k=1(k+10)!(k−1)!=∑10k=1(k+10)!⋅(k+11−(k−1))12⋅(k−1)!∑10k=1(k+11)!−(k+10)!⋅(k−1)12⋅(k−1)!==112⋅∑10k=1((k+11)!(k−1)!−(k+10)!⋅(k−1)(k−1)!)==112⋅(12!0!−11!⋅00!+13!1!−12!0!+14!2!−13!1!+...+21!9!−20!8!)=112⋅(12!0!+13!1!−12!0!+14!2!−13!1!+...+21!9!−20!8!)=21!12⋅9!ABDULAZIZABDUVALIYEV

Commented by 073 last updated on 02/Mar/23

nice solution

nicesolution

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