Question

hlo guys solve this give answer step-by-step to get a brainlist answer.​

Answer 1

Step-by-step explanation:

put n = 1

=> (1)³ - 6(1)² + 11(1) - 6

=> 12-12 = 0

=> a1=0

=> put n=2

=> (2)³ - 6(2)² + 11(2) -6

=> a2 = 8 - 24 + 22 - 6

=> a2 = 0

=> put n=3

=> a3 = (3)³-6(3)²+11(3)-6

=> a3 = 27 - 54 + 33 - 6

=> a3 = 0

=> now an = n³-6n²+11n-6

=> an = n³ - 3*2n² + 12n - n -8+2

=> an = n³ - 3*2n² + 12n-8 - n+2

=> an = n³ - 3*2n(n-2) - 2³ - n+2

=> an = (n-2)³ - (n-2)

=> when we put n≥3 value will be always positive. hence proved.

Answer 2

Answer:

hope it helps you Mark me as brainliest