Question

the nth term of a sequence is an=n^3-6n^2+11n-6.show that the first terms of this sequece are zer and the remaining term are positive.

Answer 1

n=1 an=1-6+11-6=0 n=2 a2= 8-24+22-6=0. a3= 27-54+33-6=0. a4=64-96+66-6=8 +ve qed

Answer 2

Answer:

Step-by-step explanation:an =n3  - 6n2 + 11n - 6

Now put n = 1, 2, 3

    a1 = 13 - 6*12 + 11*1 - 6

=> a1 = 1 - 6 + 11 - 6

=> a1 = 12 - 12

=> a1 = 0

a2 = 23 - 6*22 + 11*2 - 6

=> a2 = 8 - 6*4 + 22 - 6

=> a2 = 8 - 24 + 22 - 6

=> a2 = 30 - 30

=> a2 = 0

a3 = 33 - 6*32 + 11*3 - 6

=> a3 = 27 - 6*9 + 33 - 6

=> a3 = 27 - 54 + 33 - 6

=> a3 = 60 - 50

=> a3 = 0

Now put n = 4, 5

a4 = 43 - 6*42 + 11*4 - 6

=> a4 = 64 - 6*16 + 44 - 6

=> a4 = 64 - 96 + 44 - 6

=> a4 = 108 - 102

=> a4 = 6

=> a4 > 0

Again

a5 = 53 - 6*52 + 11*5 - 6

=> a5 = 125 - 6*25 + 55 - 6

=> a5 = 125 - 150 + 55 - 6

=> a5 = 180 - 156

=> a5 = 24

=> a5 > 0

and so on.

Hense, the first three terms of this sequence are zero and all other terms are positive.