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 terms are positive

Answer 1

nth term of the sequence is given by $t_n$ = (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

Hi ,

It is given that ,

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

i ) put n = 1 , in equation ( 1 ) , we get

first term = t1 = ( 1 - 1 )( 1 - 2 )( 1 - 3 )

t1 = 0 × ( - 1 ) × ( - 2 )

t1 = 0

ii ) if n = 2 ,

second term = t2 = ( 2 - 1 ) ( 2 - 2 ) ( 2 - 3 )

t2 = 1 × 0 × ( - 1 )

t2 = 0

iii ) if n = 3 ,

third term = t3 = ( 3 - 1 ) ( 3 - 2 ) ( 3 - 3 )

t3 = 2 × 1 × 0

t3 = 0

iv ) if n = 4 ,

t4 = ( 4 -1 ) ( 4 - 2 ) ( 4 - 3 )

t4 = 3 × 2 × 1

t4 = 6 > 0

v ) if n = 5 ,

t5 = ( 5 - 1 )( 5 -2 ) ( 5 - 3 )

t5 = 4 × 3 × 2

t5 = 24 > 0

Therefore ,

we conclude that ,

if n is a natural number ,

then first three terms of tn is equals to zero ,

and remaining terms are positive .

I hope this helps you.

: )