If (-1 )^r .C_r (r+1 ) (r+2 ) (r+3 )= 1a(n+b), then a + b is

Question

Accepted Answer

Correct answer: 5. (5) (-1 )^r .C_r(r+1)(r+2)(r+3) = -1(n+1)(n+2)(n+3) -1(n+1)(n+2)(n+3)( (1-1 )^n+3-( ^n+3C_0 (-1 )^0+ ^n+3C_1 (-1 )^1+ ^n+3C_2 (-1 )^2)) 1 (n+1 ) (n+2 ) (n+3 ) (n + 2) × (n+1 )2 = 12 (n+3 )

$If \sum \frac{( - 1 ) ^{r} . C _{r}}{( r + 1 ) ( r + 2 ) ( r + 3 )} = \frac{1}{a ( n + b )}, then a + b is$

Solution

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