# Prove by mathematical induction:

### Asked by ashwinikumar59 | 19th Sep, 2010, 08:28: AM

Let

f(n) = n^{7} - n

Let M(6) mean 'is a multiple of 6', and M(7)

mean 'is a multiple of 7'.

f(n) = n(n^{6} - 1)

= n(n^{3} - 1)(n^{3} + 1)

= n(n-1)(n^{2}+n+1)(n+1)(n^{2}-n+1)

= (n-1)n(n+1)(n^{2}-n+1)(n^{2}+n+1)

since we have factors (n-1)n(n+1) we have 3 consecutive numbers and

the product of ANY 3 consecutive numbers will be divisible by 2 and

also by 3, so therefore also by 6.

The problem reduces to showing that n^{7} - n = M(7). We will do that

by induction:

Show that it is true for a particular value of n, say n=2.

2^{7} - 2 = 126 = 7 x 18 = M(7)

Assume it is true for some value n=k, so we assume:

f(k) = k^{7} - k = M(7)

Now consider the next value n=k+1

We must show that f(k+1) is also M(7)

f(k+1) = (k+1)^{7} - (k+1)

= k^{7} + ^{7}C_{1}k^{6} + ..... + ^{7}C_{6}k + 1 - (k+1)

= k^{7} - k + M(7) + 1 - 1

= k^{7} - k + M(7)

k^{7} - k is divisible by 7 and so the whole expression for f(k+1) is divisible by 7.

So if f(k) is M(7) then we have shown that f(k+1) is also M(7).

Whenever f(k) is divisible by 7

Therefore n^{7} - n is always divisible by both 6 and 7 and therefore

it is divisible by 42

### Answered by | 22nd Sep, 2010, 09:44: AM

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