Question

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Answer 1

Answer:

Hølä mate answers are as follows:-☺

10).please refer to the given attachment.

11)

I) Let the given statement be P(n), i.e.,

P(n): 2^3n – 1 is divisible by 7.

It can be observed that P(n) is true for n = 1 since 2^3 × 1 – 1 = 8 – 1 = 7, which is divisible by 7.

Let P(k) be true for some positive integer k, i.e.,

P(k): 2^3k – 1 is divisible by 7.

∴2^3k – 1 = 7m; where m ∈ N … (1)

We shall now prove that P(k + 1) is true whenever P(k) is true.

Consider

Therefore, 2^3n – 1 is divisible by 7.

Thus, P(k + 1) is true whenever P(k) is true.

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

ii) Let P(n): 3 – 1 is divisible by 8, for all natural numbers n.

Now, P(l): 3 – 1 = 8, which is divisible by 8.

Hence, P(l) is true.

Let us assume that, P(n) is true for some natural number n = k.

P(k): 32k – 1 is divisible by 8

or 32k -1 = 8m, m ∈ N (i)

Now, we have to prove that P(k + 1) is true.

P(k+ 1): 32(k+1)– l

= 32k • 32 — 1

= 9(8m + 1) – 1 (using (i))

= 72m + 9 – 1

= 72m + 8

= 8(9m +1), which is divisible by 8 Thus P(k + 1) is true whenever P(k) is true.

So, by the principle of mathematical induction P(n) is true for all natural numbers n.

iii)Let p(n) be 102n-1+1 =11p

p(1) = 10+1 = 11 which is divisible by 11

Let p(k) be true.

p(k) = 102k-1+1 = 11m

p(k+1) = 102(k+1) - 1+1

=102k+2-1+1

=102k-1.102+1

=11m . 102

which is a multiple of 11.

Hence by principle of mathematical induction, p(n) is true for all nEN.

Have a nice day☺!!

Answer 2

Answer:

hii

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