(B) Let $f(n) = n(n^2-1) = (n-1)n(n+1)$. This is the product of three consecutive integers,so it is always divisible by $3! = 6$. Thus,$\alpha = 6$.Le

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(B) Let $f(n) = n(n^2-1) = (n-1)n(n+1)$. This is the product of three consecutive integers,so it is always divisible by $3! = 6$. Thus,$\alpha = 6$.Let $g(n) = 2n(n^2+2) = 2n^3 + 4n$. For $n=1$,$g(1) = 2(1)(1+2) = 6$.For $n=2$,$g(2) = 2(2)(4+2) = 24$.For $n=3$,$g(3) = 2(3)(9+2) = 66$.The greatest common divisor of these values is $\beta = 6$.Therefore,$\alpha \beta = 6 \times 6 = 36$.

Class 11 Mathematics Permutation and Combination Multinomial theorem, Number of divisors

Medium

If $\alpha$ and $\beta$ are the greatest common divisors of $n(n^2-1)$ and $2n(n^2+2)$ respectively for all $n \in N$,then $\alpha \beta=$

TS EAMCET 2018 All TS EAMCET

A
$18$
B
$36$
C
$27$
D
$9$

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Permutation and Combination

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Multinomial theorem, Number of divisors

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