If $\Gamma(n + 2) = 90\,\Gamma(n)$ ; $(n > 0)$, then the value of $'n'$ is : (a) $9$ (b) $8$ (c) $10$ (d) $7$

If Gamma(n + 2) = 90Gamma(n) ; (n > 0), then the v – Class 12 Business Mathematics Statistics

Question 12 of 42

\Gamma(n + 2) = 90\,\Gamma(n)

(a)

9

(b)

8

(c)

10

(d)

7

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Apply the Gamma-function recurrence $\Gamma(n+1) = n\,\Gamma(n)$ twice to get $\Gamma(n+2) = (n+1)n\,\Gamma(n)$ , cancel $\Gamma(n)$ , and solve $(n+1)n = 90$ .

Step 1 — Reduce $\Gamma(n+2)$ . Using $\Gamma(m+1) = m\,\Gamma(m)$ :

$\Gamma(n+2) = (n+1)\,\Gamma(n+1) = (n+1)\,n\,\Gamma(n).$

Step 2 — Substitute into the given relation.

$(n+1)\,n\,\Gamma(n) = 90\,\Gamma(n).$ …

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Q.If $\Gamma(n + 2) = 90\,\Gamma(n)$ ; $(n > 0)$ , then the value of $'n'$ is :

Concept understanding — Definite Integral (Fundamental Theorem and Properties)

Q.If Γ(n+2)=90 Γ(n)\Gamma(n + 2) = 90\,\Gamma(n)Γ(n+2)=90Γ(n) ; (n>0)(n > 0)(n>0), then the value of ′n′'n'′n′ is :

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Q.If $\Gamma(n + 2) = 90\,\Gamma(n)$ ; $(n > 0)$ , then the value of $'n'$ is :

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