第11类RD Sharma解决方案–第12章数学归纳法–练习12.1

问题1：如果P(n)是“ n(n + 1)是偶数”的陈述，那么P(3)是什么?

解决方案：

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

Substituting n with 3 we get,

P (3) = 3 (3 + 1)

P (3) = 3 (4)

P (3) = 12

Since P (3) = 12, and 12 is even

Therefore, P (3) is also even.

问题2：如果P(n)是“ n ³ + n可被3整除”的陈述，则证明P(3)是正确的，但P(4)是不正确的。

解决方案：

Given: P (n) = n³ + n is divisible by 3

Firstly, substituting n with 3 we get,

P (3) = 3³ + 3

P (3) = 27 + 3

P (3) = 30

Since P (3) = 30, and 30 is divisible by 3

Therefore, P (3) is true.

Now, substituting n with 4 we get,

P (4) = 4³ + 4

P (4) = 64 + 4

P (4) = 68

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

问题3：如果P(n)是语句“2 ^n个≥3N”，如果P(r)为真，证明P(R + 1)为真。

解决方案：

Given: P (n) = 2ⁿ ≥ 3n and p(r) is true.

also given that P (r) is true

When we substitute n with r we get

P (r) = 2^r ≥ 3r

Now, multiply both sides by 2 we get,

2 × 2^r ≥ 3r × 2

2^{r + 1} ≥ 6r

We can write 6r as 3r + 3r

2^{r + 1} ≥ 3r + 3r

Since 3^r ≥ 3, therefore 3r + 3r ≥ 3 + 3r

Substituting  3r + 3r with 3 + 3r we get,

2^{r + 1} ≥ 3 + 3r

2^{r + 1} ≥ 3(r + 1)      [Taking 3 as common]

Since, 2^r+1 ≥ 3(r + 1) is equal to P (r + 1)

Therefore, P (r + 1) is true.

问题4：如果P(n)是语句“ n ² + n为偶数”，并且如果P(r)为真，则P(r + 1)为真

解决方案：

Given: P (n) = n² + n is even 

Also given that P (r) is true,

Therefore, P (r) = r² + r is even

Let us consider r² + r = 2x … (i)

Substituting r with r + 1

Now, (r + 1)² + (r + 1)

r² + 1 + 2r + r + 1    [ formula = (a + b)² = a² + 2ab + b²]

(r² + r) + 2r + 2

2x + 2r + 2 [from equation (i) we get 2x = r² + r]

2(x + r + 1)

2μ

Since, (r + 1)² + (r + 1) is Even which is equal to P (r + 1).

Therefore, P (r + 1) is true.

问题5：给出一个陈述P(n)的示例，使得对所有n ϵ N都成立。

解决方案：

Let us consider P (n) as

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

Since P (n) is true for all natural numbers.

Therefore, P (n) is true for all n ∈ N.

问题6：如果P(n)是“ n ² – n + 41是素数”的陈述，请证明P(1)，P(2)和P(3)是正确的。还证明P(41)不正确。

解决方案：

Given: P(n) = n² – n + 41 is prime.

Substituting n with 1, we get

P (1) = 1 – 1 + 41

P (1) = 41

Since P (1) = 41, and 41 is prime. 

Therefore, P (1) is true.

Now substituting n with 2, we get

P(2) = 2² – 2 + 41

P(2) = 4 – 2 + 41

P(2) = 43

Since P (2) = 43, and 43 is prime.

Therefore, P (2) is true.

Now substituting n with 3, we get

P (3) = 3² – 3 + 41

P (3) = 9 – 3 + 41

P (3) = 47

Since P (3) = 47, and 47 is prime.

Therefore, P (3) is true.

Now substituting n with 41, we get

P (41) = (41)² – 41 + 41

P (41) = 1681

Since P (41) = 1681, and 1681 is not prime.

Therefore, P (41) is not true.