Question

for any positive integer n prove that nq-n is divisible by 6

Answer 1

hey dear


here is your answer


Solution


Given


S(n) = n^3 -n divisible by 6


Let N =1 then we get 0

which is divisible by 6


Let us assume that n = K

S (K) = K^3 - k

which is divisible by 6

S(k) is true


(k^3 - K) = 6 = m. ( m is integer)

( K ^3 - K) = 6m

k^3 = 6m + K


Now we have to prove that n = K +1

(K + 1)^3 - (K +1)

= (K^3 + 3k^2 +3k +1) - (k+1)

substitute equation 1 in above equation

6m + K + 3k ^2 +2k

6m + 3 k^2 +k

6m + 3k (k+1) ( K ( k+1)

( 2p is an even number and p is a natural no)

6m + 3*2p

6m + 6mp

6 ( m+p) ( 6 taken common)

So S ( k+1) is true


Hence 6 divisible by n^3 -n


Hence proved


hope it helps

thank you