TS EAMCET · Maths · Binomial Theorem
\(n^5-5 n^3+4 n\) is divisible by 120 is true for
- A all positive integers \(n\)
- B all positive integers for \(n \geq 3\) only
- C all positive integers for \(n \geq 1\) only
- D all positive integers for \(n \geq 5\) only
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…
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