Saved Bookmarks
| 1. |
n(n+1)(n+5) is a multiple of 3. |
|
Answer» When n = 1, n(n+1)(n+5) = 1*2*6 = 12, which is divisible by 3. So let he statement P(K): k(k+1)(k+5) is divisible by 3 is TRUE. ii) P(k+1): (k+1)(k+2)(k+6) ==> P(k+1) = (k+1)(k+2){(k+5) + 1} = [k(k+1) + 2(k+1)]*{(k+5) + 1} = k(k+1)(k+5) + 2(k+1)(k+5) + k(k+1) + 2(k+1) = k(k+1)(k+5) + (k+1)[2(k+5) + k + 2] = k(k+1)(k+5) + (k+1)(3k+12) = k(k+1)(k+5) + 3(k+1)(k+4) Thus, P(k+1): P(k) + Multiple of 3 [From (i) P(k) is k(K+1)(k+5) is divisible by 3] So both terms are divisible by 3 and hence P(k+1) is also divisible by 3 Hence it is proved that the statement is divisible by for two consecutive terms. So the statement "n(n+1)(n+5) is divisible by 3 for all n element of natural numbers" is TRUE. |
|