જો ${1 \over {x(x + 1)\,(x + 2)....(x + n)}} = {{{A_0}} \over x} + {{{A_1}} \over {x + 1}} + {{{A_2}} \over {x + 2}} + .... + {{{A_n}} \over {x + n}}$ તો ${A_r} = $
Easy
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b
(b) \(1 = {A_0}(x + 1)\,(x + 2)....(x + n) + {A_1}x(x + 2)\,(x + 3)...(x + n)\)
(b) \(1 = {A_0}(x + 1)\,(x + 2)....(x + n) + {A_1}x(x + 2)\,(x + 3)...(x + n)\)
\( + ... + {A_r}x(x + 1)\,(x + 2)....(x + r - 1)\,(x + r + 1)\,(x + r + 2)\) \(.......(x + n)\)
Putting \(x = - r\),
\(1 = {A_r}( - r)\,( - r + 1)\,( - r + 2),\,.....( - 1).1.2....( - r + n)\)
\( \Rightarrow \) \(1 = {A_r}.{( - 1)^r}r!.(n - r)\,!\); \({A_r} = {{{{( - 1)}^r}} \over {r\,!\,(n - r)\,!}}\).
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