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If ((3n+1)!)/(3n−1)! = 156, what is the value of n?

Answer and explanation

Correct answer: 4

Answer: 4, so option B is correct. Expand the numerator only until the denominator appears: (3n+1)!=(3n+1)(3n)(3n−1)!. After cancelling (3n−1)!, the equation becomes (3n+1)(3n)=156. Put t=3n. Then t(t+1)=156. The consecutive integers 12 and 13 have product 12×13=156, so t=12 and 3n=12. Dividing by 3 gives n=4. Verification: for n=4, the original expression is 13!/11!=13×12=156. Option A gives 10×9=90, option C gives 16×15=240, and option D gives 19×18=342. Thus only B works. The key is to identify the two consecutive factors rather than expand all factorials.

Related tags

Factorial NotationFactorial CancellationConsecutive IntegersAlgebraic EquationMathematicsClass 11 McqPermutations And Combinations

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