(n+2)!=2550 × n!.
Find n?
Answers
Answered by
61
we know n! =n (n-1) (n-2)(n-3).......1
now according to question
(n+2)(n+1)n!=2550n!
now
(n+2)(n+1)=2550
n^2+3n+2=2550
n^2+3n-2548=0
n=-52,49
but for factorial n> 0
so n=49
now according to question
(n+2)(n+1)n!=2550n!
now
(n+2)(n+1)=2550
n^2+3n+2=2550
n^2+3n-2548=0
n=-52,49
but for factorial n> 0
so n=49
abhi178:
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Answered by
32
(n+2)!=2250(n!)
(n+2)(n+1)(n!)=2550(n!)
(n+2)(n+1)=2550
n^2+3n+2-2550=0
n^2+3n-2548=0
n(n+52)-49(n+52)=0
(n-49)(n+52)=0
n=49
n=-52
as n cannot be negetive so n=49
(n+2)(n+1)(n!)=2550(n!)
(n+2)(n+1)=2550
n^2+3n+2-2550=0
n^2+3n-2548=0
n(n+52)-49(n+52)=0
(n-49)(n+52)=0
n=49
n=-52
as n cannot be negetive so n=49
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