Question

If the nth term of a sequence be tn=(n-1) (n-2) (n-3), then show that the first three terms of the sequence are zero, but the rest of the terms are positive.

Answer 1

nth term of the sequence is given by tn = (n - 1)(n -2 )(n -3 ), here n is integers e.g., n = 1, 2, 3, 4, .........

First term , put n = 1
Then, t₁ = (1 - 1)(1 - 2)(1 - 3) = 0 × (-1) × (-2) = 0

2nd term, put n = 2
Then, t₂ = (2 - 1)(2 - 2)(2 - 3) = 1 × 0 × (-1) = 0

3rd term , put n = 3
Then, t₃ = (3 - 1)( 3 - 2)(3 - 3) = 2 × 1 × 0 = 0

Here it is clear that first three terms of the sequence are zero.

Now, put n = 4
Then, t₄ = (4 - 1)(4 - 2)(4 - 3) = 3 × 2 × 1 = 6 > 0 e.g., positive
Put n = 5
Then, t₅ = (5 - 1)(5 - 2)(5 - 3) = 4 × 3 × 2 = 24 > 0 e.g., positive
........... ...... .....
...................................
Hence, after 3rd term all the terms of the sequence are positive.

Answer 2

Hello Dear.

Here is the answer---


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 Given, nth term of the sequence is given by the---

            tn = (n - 1)(n - 2)(n - 3)
  
Proof -) 

For n = 1
 
          T₁ = (1 - 1)(1 - 2)(1 - 3)
              = 0 × -1 × -2
              = 0

For n = 2

        T₂ = ( 2 - 1)(2 - 2)(2 - 3)
            = 1 × 0 × -1
            = 0

For n = 3

       T₃ = (3 - 1)(3 - 2)(3 - 3)
            = 2 × 1 × 0 
            = 0

For n = 4

     T₄ = (4 - 1)(4 - 2)(4 - 3)
          = 3 × 2 × 1
          = 6

From the above calculations, we can observe that the First three terms are zero but the fourth term is 6 which is positive.

Hence Proved.


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Have a Marvelous Day.