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TS EAMCET · Maths · Binomial Theorem

\(n^5-5 n^3+4 n\) is divisible by 120 is true for

  1. A all positive integers \(n\)
  2. B all positive integers for \(n \geq 3\) only
  3. C all positive integers for \(n \geq 1\) only
  4. D all positive integers for \(n \geq 5\) only
Verified Solution

Answer & Solution

Correct Answer

(A) all positive integers \(n\)

Step-by-step Solution

Detailed explanation

Since, \(n^5-5 n^3+4 n=n\left(n^4-5 n^2+4\right)\) \(=n\left(n^4-4 n^2-n^2+4\right)=n\left(n^2-1\right)\left(n^2-4\right)\) \(=(n-2)(n-1)(n)(n+1)(n+2)=P(n)\) (let) For \(n=1,2, P(n)\) is zero so it is divisible by 120 . and for \(n=3, P(3)=5 !=120\) and so on, so the \(P(n)\) is…