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Find n, ""^(n+5)P(n+1)=(11)/(2)(n-1)*""...

Find n, ` ""^(n+5)P_(n+1)=(11)/(2)(n-1)*""^(n+3)P_(n).`

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We have, `(.^(n+5)P_(n+1))/(.^(n+3)P_(n))=(11(n-1))/(2)`
`implies((n+5)(n+3)^(n+3)P_(n-1))/(.^(n+3)P_(n))=(11(n-1))/(2)` [from note (iii)]
`((n+5)(n+4))/((n+3-n+1))=(11(n-1))/(2)` [from note (v)]
`implies(n+5)(n+4)=22(n-1)`
`impliesn^(2)-13n+42=0`
`implies(n-6)(n-7)=0`
`thereforen=6.7`
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