Arun
Last Activity: 7 Years ago
let P(n)= n(n+ 1)(n + 5)
P(1) = 1* 2* 6 = 12, divisible by 3
let P(k) be true
P(k) = k * (k +1) *(k +5)
on solving
P(k) = k^3 + 6k^2 + 5k
= 3 y (multiple of 3)
k^3 = 3 y – 6k^2 –5k …........(1)
now
P(k + 1) = (k + 1) (k + 1+ 1) (k + 1+ 5)
= (k + 1) (k + 2) (k + 6)
on solving
= k^3 + 9k^2 + 20 k +12
put value of k^3 from eqn (1)
= 3y –6k^2 – 5k +9k^2 + 20k + 12
= 3y + 3k^2 +15 k +12
= 3(y + k^2 + 5k +4)
therefore we can say P(k + 1) is a multiple of 3.
hope it helps