We can write the given statement as
P (n): 102n – 1 + 1 is divisible by 11
If n = 1 we get
P (1) = 102.1 – 1 + 1 = 11, which is divisible by 11
Which is true.
Consider P (k) be true for some positive integer k
102k – 1 + 1 is divisible by 11
102k – 1 + 1 = 11m, where m ∈ N …… (1)
Now let us prove that P (k + 1) is true.
Here
10 2 (k + 1) – 1 + 1
We can write it as
= 10 2k + 2 – 1 + 1
= 10 2k + 1 + 1
By addition and subtraction of 1
= 10 2 (102k-1 + 1 – 1) + 1
We get
= 10 2 (102k-1 + 1) – 102 + 1
Using equation 1 we get
= 102. 11m – 100 + 1
= 100 × 11m – 99
Taking out the common terms
= 11 (100m – 9)
= 11 r, where r = (100m – 9) is some natural number
10 2(k + 1) – 1 + 1 is divisible by 11
P (k + 1) is true whenever P (k) is true.
Therefore, by the principle of mathematical induction, statement P (n) is true for all natural numbers i.e. n.
Answered by Abhisek | 1 year agoGiven an example of a statement P (n) such that it is true for all n ϵ N.
a + (a + d) + (a + 2d) + … + (a + (n-1)d) = \( \dfrac{n}{2}\) [2a + (n-1)d]
72n + 23n – 3. 3n – 1 is divisible by 25 for all n ϵ N
n (n + 1) (n + 5) is a multiple of 3 for all n ϵ N.