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If sum(t=1)^(1003) (r^(2) + 1)r! = a! - ...

If `sum_(t=1)^(1003) (r^(2) + 1)r! = a! - b(c!)` where `a, b, c in N` the least value of `(a + b + c)` is `pqrs` then

A

`p + q+ r + s = 4`

B

`(p + q)/(r + s) = 1`

C

`(p + q + r)/(s) = 3`

D

`p.q.r.s` is even

Text Solution

Verified by Experts

The correct Answer is:
A, B, C, D
`sum_(f=1)^(1003)(r^(2) + 1) r! = sum_(r=1)^(1003)((r^(2) + r)-(r - 1))rt`
`= sum_(f=1)^(1003)(r(r + 1)! - (r - 1)r!)`
`= (1.2! - 0) + (2.3! - 1.2!) + (3.4! - 2.3!)+ …….`
`= 1005! - 2(1004!) = a! - b(c!)`
`:. (a + b + c)_("least") = 1005 + 2 + 1004 = 2011 = pqrs`
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