Show that `n^7-n` is divisible by 42 .

Explanation

Let `P(n)=n^7-n`
Step I For `n=1`.
`P(1)=1^7-=0` , which is divisible by 42.
Therefore , the result is true for `n=1` .
Step II Assume that the result is true for `n=k`. Then ,
`P(k)=k^7-k` is divisible by 42.
`rArr P(k)=42r`, where r is an integer.
Step III For `n=k+1`.
`P(k+1)=(k+1)^7-(k+1)=(1+k)^7-(k+1)`
`=1+.^(7)C_(1)k+.^(7)C_(2)k^2+.^(7)C_(3)k^3+.^(7)C_(4)k^4+.^(7)C_(5)k^5+.^(7)C_(6)k^6+.^(7)C_(7)k^7-(k+1)`
`=(k^7-k)+(.^7C_1k+.^7C_2k^2+.^7C_(3)k^3+.^7C_4k^4+.^7C_5+.^(7)C_(6)k^6)`
But by assumption `k^7-k` is divisible by 42.
Also `.^7C_1k+.^7C_2k^2+.^7C_3k^3+.^7C_4k^4+.^7C_5k^5+.^7C_6k^6` is divisible by 42. `[ because .^7C_r,1 le r le 6 "is divisible by" 7]`
Hence , `P(k+1)` is divisible by 42. This shows that , the result is true for `n=k+1`.
`therefore` By the principle of mathematical induction , the result is true for all `n in N`.