1.

(1)/(x(x - 1 ) (x + 2 ) …. (x+ n))= (A_0 )/( x )+(A _ 1 )/(x + 1 )+…….. ( A _ n) /(x +n ) ,0leile r rArr A _ ris equalto

Answer»

` ( -1) ^ r(r!)/( (n - r) !) `
` (-1) ^r (r!)/(r!(n-r )!) `
` (1)/(r!(n-r)!)`
`(r!)/((n-r)!)`

Solution : ` (1)/(x (x +1)(x+2)… (x+n))= (A_0)/(x)+ (A_1)/((x+1)) + … + (A_n)/(x+n) ,0le ILE r `
` rArr(1)/(x(x+1)(x+2)…(x+r)…(x+n)) = (A_0)/(x)+ (A_1)/((x+1))+ …. + (A_n)/(x+n ) `
` (1)/(x(x+1)(x+2)… (x+r - 1 )(x + r + 1 )… (x + n))`=
`rArr(A_0 (x + r ))/(x)+ (A_1(x+r))/(x+1) + ... + (A_r(x+r))/((x+r)) +(A_(r +1)(x+r))/(x + r + 1 )+... + (A_n(x+r))/(x + n) `
` rArr (1)/(x(x+1)(x+2)...(x+n)) = (A_0 (x + r ))/(x)+ (A_1(x+r))/(x + 1 )+... + A_r + A _ (r + 1 )((x + r ))/((x+r + 1)) + ... + A_n((x + r ))/((x + n)) `
Substituting`x =- r `in ABOVEEQUATION, we GET,
` rArr(1)/((-r)(- r + 1 ) (-r + 2 )... (-r + r - 1 )(-r + r + 1 )... (-r + n)) `
`= 0 + 0+ ...+ A _ r + 0 `
`thereforeA_r= (1 ) /((-1) ^r r (r -1) ... (-1).(1) (2) ... (-r + n)) `
` A _r =((-1) ^r)/((r) (r - 1 )...(1).(1)(2)...(n-r )) `
`= ((-1) ^r ) /(r! (n - r) ! ) `


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