If n -1 -12 - n -1 - then the value of n can beA- -4B- 3C- -3D- 4

The correct options are B 3(n+1)! = 12× (n 1)! ⇒ (n+1)n(n 1)! = 12× (n 1)! ⇒ n^2+n = 12 (∵ n 1 0) ⇒ n^2+n 12 = 0 ⇒ (n+4)(n 3) = 0 ⇒ n = 4, 3 ∵ n= 4 does n ...

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Standard XII

Mathematics

Pascal Triangle

If n +1 !=12 ...

Question

If $(n + 1)! = 12 \times (n - 1)!$ , then the value of $n$ can be

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Solution

The correct option is B $3$
$(n + 1)! = 12 \times (n - 1)! \Rightarrow (n + 1) n (n - 1)! = 12 \times (n - 1)! \Rightarrow n^{2} + n = 12 (∵ n - 1 \neq 0) \Rightarrow n^{2} + n - 12 = 0 \Rightarrow (n + 4) (n - 3) = 0 \Rightarrow n = - 4, 3$
$∵ n = - 4$ does not satisfy the given equation.
$∴ n = 3, (∵ n \in N)$

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If (n+1)!=12×(n−1)!, then the value of n can be

The correct option is B 3(n+1)!=12×(n−1)!⇒(n+1)n(n−1)!=12×(n−1)!⇒n2+n=12 (∵n−1≠0)⇒n2+n−12=0⇒(n+4)(n−3)=0⇒n=−4,3 ∵n=−4 does not satisfy the given equation. ∴n=3, (∵n∈N)

If $(n + 1)! = 12 \times (n - 1)!$ , then the value of $n$ can be