prove that (10^n) - 1 is divisible by 11 for every EVEN natural number n (by induction)?

Let n = 2. Then 10^2 - 1 = 100 - 1 = 99 = 11*9 so 11 | 99. This holds for n = 2.

Assume this holds for some n = 2k. Need to show this holds for n = 2(k+1) = 2k + 2.

10^(2k + 2) - 1

10^2 * 10^2k - 1

100 * 10^2k - 1

(99 + 1)(10^2k) - 1

99 * 10^2k + 10^2k - 1

Clearly, 11 | 99 * 10^2k since 11 | 99

By induction assumption, 11 | (10^2k - 1)

Thus, 11 | (99 * 10^2k + 10^2k - 1) i.e. 11 | (10^(2k+2) - 1)

This completes the proof.

Base case: n = 2

100 - 1 = 99 = 9*11

Assume true for all even n.

Test first even n+2.

10^(n+2) - 1

10²*10ⁿ - 1

By the induction hypothesis, as n < n+2 and 2 < n+2 and 2 and n are even,

10² - 1 and 10ⁿ - 1 are both divisible by 11, but that means 10² and 10ⁿ have remainder 1 when divided by 11, i.e. 10² = 1 (mod 11) and 10ⁿ = 1 (mod 11). Therefore 10²*10ⁿ = 1*1 (mod 11) = 1 (mod 11), so that by the definition of congruences 10²*10ⁿ - 1 = 10^(n+2) - 1 is divisible by 11 and we're done.

It is proved that for n= 2, 4,6 8 , it is divisible by 11

Let us assume that it is divisible by 11 for n = m , if we can prove it

it is divisible by 11 for n= m+2 , then it is divisible by 11 for all even n

T(m) = (10^m) - 1 for n= m

for n= m+2 , T(m+2) = 10 ^ (m+2) - 1 = 10 ^m * 100 - ( 100 -99)

= 100 ( 10 ^m - 1) + 11 * 9 = 100 * T(m) + 11*9 , T(m) is already divisible by 11

Hence T(m+2) is divisible by 11 .

A number is divisible by 11 if the alternating sum of its decimal digits is 0. The number 10^n - 1 is represented in decimal form by n consecutive 9's. If n is even then the alternating sum of digits is

9 - 9 + 9 - 9 + ... = 0.

So, all we need to do is show that n is even for every even natural number n by induction.

If n = 2, 2/2 = 1 which is natural so this is true for n = 2.

Assume it is true for some n = k where k is an even natural number that k/2 is an integer.

For n = k + 2, we have (k+2)/2 = k/2 + 2/2 which, by the induction hypothesis and the base case, is the sum of two integers and is, therefore an integer.

Therefore, by proof by mathematical induction, it is true that every even natural number n is even.

;)