**Answer
1** :

Given:

P (n) = n (n + 1) is even.

So,

P (3) = 3 (3 + 1)

= 3 (4)

= 12

Hence, P (3) = 12, P (3) is also even.

If P (n) is the statement “n^{3} +n is divisible by 3”, prove that P (3) is true but P (4) is not true.

**Answer
2** :

Given:

P (n) = n^{3} + n is divisible by 3

We have P (n) = n^{3} + n

So,

P (3) = 3^{3} + 3

= 27 + 3

= 30

P (3) = 30, So it is divisible by 3

Now, let’s check with P (4)

P (4) = 4^{3} + 4

= 64 + 4

= 68

P (4) = 68, so it is not divisible by 3

Hence, P (3) is true and P (4) is not true.

If P (n) is the statement “2^{n} ≥3n”, and if P (r) is true, prove that P (r + 1) is true.

**Answer
3** :

Given:

P (n) = “2^{n} ≥ 3n” and p(r) is true.

We have, P (n) = 2^{n} ≥ 3n

Since, P (r) is true

So,

2^{r}≥ 3r

Now, let’s multiply both sides by 2

2×2^{r}≥ 3r×2

2^{r + 1}≥ 6r

2^{r + 1}≥ 3r + 3r [since 3r>3 = 3r + 3r≥3 +3r]

∴ 2^{r + 1}≥ 3(r + 1)

Hence, P (r + 1) is true.

**Answer
4** :

Given:

P (n) = n^{2} + n is even and P (r) is true,then r^{2} + r is even

Let us consider r^{2} + r = 2k … (i)

Now, (r + 1)^{2} + (r + 1)

r^{2} + 1 + 2r + r + 1

(r^{2} + r) + 2r + 2

2k + 2r + 2 [from equation (i)]

2(k + r + 1)

2μ

∴ (r + 1)^{2} + (r + 1) is Even.

Hence, P (r + 1) is true.

**Answer
5** :

Let us consider

P (n) = 1 + 2 + 3 + – – – – – + n = n(n+1)/2

So,

P (n) is true for all natural numbers.

Hence, P (n) is true for all n ∈ N.

**Answer
6** :

Given:

P(n) = n^{2} – n + 41 is prime.

P(n) = n^{2} – n + 41

P (1) = 1 – 1 + 41

= 41

P (1) is Prime.

Similarly,

P(2) = 2^{2} – 2 + 41

= 4 – 2 + 41

= 43

P (2) is prime.

Similarly,

P (3) = 3^{2} – 3 + 41

= 9 – 3 + 41

= 47

P (3) is prime

Now,

P (41) = (41)^{2} – 41 + 41

= 1681

P (41) is not prime

Hence, P (1), P(2), P (3) are true but P (41) is not true.

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