第 11 课 RD Sharma 解决方案 - 第 12 章数学归纳 - 练习 12.2 |设置 2

用数学归纳原理证明如下：

问题 17. a + ar + ar ² + … + ar ^n-1 = a [(r ⁿ – 1)/(r – 1)], r ≠ 1

解决方案：

Let P (n) = a + ar + ar² + … + ar^n-1 = a [(rⁿ – 1)/(r – 1)]

Step 1:

Now, let us check P(n) for n = 1.

P (1) = a = a (r¹ – 1)/(r – 1)

or, a = a

So, P(1) is true.

Step 2:

Let us consider P (n) be the true for n = k,

P (k) = a + ar + ar2 + … + ar^k-1 = a [(r^k – 1)/(r – 1)]     … (i)

Step 3:

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

Then, a + ar + ar² + … + ar^k-1 + ar^k

Now, putting the value (i), we get,

= a [(r^k – 1)/(r – 1)] + ar^k              (using eq(i))

= a[r^k – 1 + r^k(r – 1)] / (r – 1)

= a[r^k – 1 + r^k+1 – r^-k] / (r – 1)

= a[r^k+1 – 1] / (r – 1)

So, P (n) is true for n = k + 1

i.e., P (n) is true for all n ∈ N

Hence, by principle of Mathematical Induction (PMI), P (n) is true for all n ∈ N.

问题 18. a + (a + d) + (a + 2d) + … + (a + (n – 1)d) = n/2 [2a + (n – 1)d]

解决方案：

Let P (n) = a + (a + d) + (a + 2d) + … + (a + (n – 1)d) = n/2 [2a + (n – 1)d]

Step 1:

Now, let us check P(n) for n = 1.

P (1) = a = 1/2[2a + (1 – 1)d]

or, a = a

So, P(1) is true.

Step 2:

Let us consider P (n) be the true for n = k,

P (k) = a + (a + d) + (a + 2d) + … + (a + (k – 1)d) = k/2 [2a + (k – 1)d]    … (i)

Step 3:

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

Then, a + (a + d) + (a + 2d) + … + (a + (k – 1)d) + (a + (k)d)

Now, putting the value eq(i), we get,

= k/2 [2a + (k – 1)d] + (a + kd)               (Using eq(i))

= [2ka + k(k – 1)d + 2(a + kd)] / 2

= [2ka + k2d – kd + 2a + 2kd] / 2

= [2ka + 2a + k2d + kd] / 2

= [2a(k + 1) + d(k2 + k)] / 2

= (k + 1)/2 [2a + kd]

So, P (n) is true for n = k + 1

i.e., P (n) is true for all n ∈ N

Hence, by principle of Mathematical Induction (PMI), P (n) is true for all n ∈ N.

问题 19. 5 ²ⁿ – 1 可被 24 整除，对于所有 n ∈ N

解决方案：

Let P(n) = 5²ⁿ – 1 is divisible by 24

Step 1: 

Now, let us check P(n) for n = 1.

P (1) = 5² – 1 = 25 – 1 = 24

So, P(1) is true as P(n) is divisible by 24.

Step 2:

Let us consider P (n) be the true for n = k,

P (k) = 5^2k – 1 is divisible by 24

or, 5^2k – 1 = 24λ   … (i)

Step 3:

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

i.e., 5^2(k+1) – 1 is divisible by 24

Or, 5^2(k+1) – 1 = 24μ

Then, 5^{2(k + 1)} – 1

= 5^2k.5² – 1

= 25.5^2k – 1

Now, putting the value eq(i), we get,

= 25.(24λ + 1) – 1           (Using eq(i))

= 25.24λ + 24

= 24λ

So, P (n) is true for n = k + 1

i.e., P (n) is true for all n ∈ N

Hence, by principle of Mathematical Induction (PMI), P (n) is true for all n ∈ N.

问题 20. 3 ²ⁿ + 7 可以被 8 整除 对于所有 n ∈ N

解决方案：

Let P (n) =   3²ⁿ + 7 is divisible by 8

Step 1:

Now, let us check P(n) for n = 1.

P (1) = 32 + 7 = 9 + 7 = 16

So, P(1) is true as P(n) is divisible by 8.

Step 2:

Let us consider P (n) be the true for n = k,

P (k) = 3^2k + 7 is divisible by 8

or, 3^2k + 7 = 8λ

or, 3^2k = 8λ – 7   … (i)

Step 3:

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

i.e., 3^2(k+1) + 7 is divisible by 8

or, 3^2k+2 + 7 = 8μ

Then,

= 3^2k+2 + 7

= 3^2k.3² + 7

= 9.3^2k + 7

Now, putting the value eq(i), we get,

= 9.(8λ – 7) + 7                     (Using eq(i))

= 72λ – 63 + 7

= 72λ – 56

= 8(9λ – 7)

= 8μ

So, P (n) is true for n = k + 1

i.e., P (n) is true for all n ∈ N

Hence, by principle of Mathematical Induction (PMI), P (n) is true for all n ∈ N.

问题 21. 5 ²ⁿ⁺² – 24n – 25 可被 576 整除 对于所有 n ∈ N

解决方案：

Let P (n) = 5²ⁿ⁺² – 24n – 25 is divisible by 576

Step 1:

Now, let us check P(n) for n=1.

P (1) = 5^{2.1 + 2} – 24.1 – 25 = 625 – 49 = 576

So, P(1) is true as P (n) is divisible by 576.

Step 2:

Let us consider P (n) be the true for n = k,

P (k) = 5^2k+2 – 24k – 25 is divisible by 576

or, 5^2k+2 – 24k – 25 = 576λ  … (i)

Step 3:

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

i.e., 5^{2k + 4} – 24(k + 1) – 25 is divisible by 576

or, 5^{2k + 2} – 24k – 25 = 576λ

Then,

= 5^{(2k + 2) + 2} – 24(k + 1) – 25

= 5^2k+2.5² – 24k – 24 – 25

Now, putting the value eq(i), we get,

= (576λ + 24k + 25)25 – 24k – 49                   (using eq(i))

= 25. 576λ + 576k + 576

= 576(25λ + k + 1)

= 576μ

So, P (n) is true for n = k + 1

i.e., P (n) is true for all n ∈ N

Hence, by principle of Mathematical Induction (PMI), P (n) is true for all n ∈ N.

问题 22. 3 ^{2n + 2} – 8n – 9 可被 8 整除 对于所有 n ∈ N

解决方案：

Let P (n) = 3^{2n + 2}  – 8n – 9 is divisible by 8

Step 1:

Now, let us check P(n) for n = 1.

P (1) = 3^{2.1 + 2} – 8.1 – 9 = 81 – 17 = 64

So, P(1) is true as P (n) is divisible by 8.

Step 2:

Let us consider P (n) be the true for n = k,

P (k) = 3^{2k + 2} – 8k – 9 is divisible by 8

or, 3^{2k + 2} – 8k – 9 = 8λ   … (i)

Step 3:

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

i.e., 3^{2k + 4} – 8(k + 1) – 9 is divisible by 8

3^{(2k + 2) + 2} – 8(k + 1) – 9 = 8μ

Then,

= 3^2(k+1).3² – 8(k + 1) – 9

= (8λ + 8k + 9)9 – 8k – 8 – 9

Now, putting the value (i), we get,

= 72λ + 72k + 81 – 8k – 17              (using eq(i))

= 8(9λ + 8k + 8)

= 8μ

So, P (n) is true for n = k + 1

i.e., P (n) is true for all n ∈ N

Hence, by principle of Mathematical Induction (PMI), P (n) is true for all n ∈ N.

问题 23. (ab) ⁿ = a ⁿ b ⁿ对于所有 n ∈ N

解决方案：

Let P (n) = (ab)ⁿ = aⁿ bⁿ 

Step 1:

Now, let us check P(n) for n = 1.

P (1) = ab = ab

So, P(1) is true.

Step 2:

Let us consider P (n) be the true for n = k,

P (k) = (ab)^k = a^k b^k   … (i)

Step 3:

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

i.e., (ab)^k+1 = a^k+1.b^k+1

Then,

= (ab)^k+1

= (ab)^k .(ab)

Now, putting the value eq(i), we get,

= (a^k b^k) (ab)                (using eq(i))

= (a^k+1) (b^k+1)

So, P (n) is true for n = k + 1

i.e., P (n) is true for all n ∈ N

Hence, by principle of Mathematical Induction (PMI), P (n) is true for all n ∈ N.

问题 24. 对于所有 n ∈ N，n (n + 1) (n + 5) 是 3 的倍数。

解决方案：-

Let P (n) = n (n + 1) (n + 5) is a multiple of 3

Step 1: 

Now, let us check P(n) for n = 1.

P (1) = 1 (1 + 1) (1 + 5)

or, 2 × 6

or, 12

So, P(1) is true as P(n) is multiple of 3.

Step 2:

Let us consider P (n) be the true for n = k,

P (k) = k (k + 1) (k + 5) is a multiple of 3

or, k(k + 1) (k + 5) = 3λ   … (i)

Step 3:

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

i.e., (k + 1)[(k + 1) + 1][(k + 1) + 5] is a multiple of 3

or, (k + 1)[(k + 1) + 1][(k + 1) + 5] = 3μ

Then,

 (k + 1) [(k + 1) + 1] [(k + 1) + 5]

= (k + 1) (k + 2) [(k + 1) + 5]

= [k (k + 1) + 2(k + 1)] [(k + 5) + 1]

= k (k + 1) (k + 5) + k(k + 1) + 2(k + 1) (k + 5) + 2(k + 1)

= 3λ + k2 + k + 2(k2 + 6k + 5) + 2k + 2

= 3λ + k2 + k + 2k2 + 12k + 10 + 2k + 2

= 3λ + 3k2 + 15k + 12

= 3(λ + k2 + 5k + 4)

= 3μ

So, P (n) is true for n = k + 1

i.e., P (n) is true for all n ∈ N

Hence, by principle of Mathematical Induction (PMI), P (n) is true for all n ∈ N.

问题 25. 7 ²ⁿ + 2 ^3n-3 .3 ^n-1可以被 25 整除 对于所有 n ∈ N

解决方案：

Let P (n) = 7²ⁿ + 2^3n-3 . 3^n-1 is divisible by 25

Step 1:

Now, let us check P(n) for n = 1.

P (1) = 7² + 2⁰.3⁰

or, 49 + 1 = 50

So, P(1) is true as P (n) is divisible by 25.

Step 2:

Let us consider P (n) be the true for n = k,

P (k) = 7^2k + 2^3k-3. 3^k-1 is divisible by 25

or, 7^2k + 2^3k-3. 3^k-1 = 25λ   … (i)

Step 3:

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

i.e., 7^2(k+1) + 2^3k. 3^k is divisible by 25

or, 7^2k+2 + 2^3k. 3^k = 25μ

Then,  

7^2(k+1) + 2^3k. 3^k

= 7^2k.7² + 2^3k. 3^k

Now, putting the value (i), we get,

= (25λ – 2^3k-3. 3^k-1) 49 + 2^3k. 3^k                 (Using eq(i))

= 25λ. 49 – 2^3k/8. 3^k/3. 49 + 2^3k. 3^k

= 24×25×49λ – 2^3k . 3^k . 49 + 24 . 2^3k.3^k

= 24×25×49λ – 25 . 2^3k. 3^k

= 25(24 . 49λ – 2^3k. 3^k)

= 25μ

So, P (n) is true for n = k + 1

i.e., P (n) is true for all n ∈ N

Hence, by principle of Mathematical Induction (PMI), P (n) is true for all n ∈ N.

问题 26. 2.7 ⁿ + 3.5 ⁿ – 5 可被 24 整除，对于所有 n ∈ N

解决方案：-

Let P (n) = 2.7ⁿ + 3.5ⁿ – 5 is divisible by 24

Step 1:

Now, let us check P(n) for n=1.

P (1) = 14 + 15 – 5 = 24

So, P(1) is true as P (n) is divisible by 24.

Step 2:

Let us consider P (n) be the true for n = k,

P (k) = 2.7^k + 3.5^k – 5 is divisible by 24

or, 2.7^k + 3.5^k – 5 = 24q  … (i)

Step 3:

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

i.e., 2.7^k+1 + 3.5^k+1 – 5

⇒ 2.7.7^k + 3.5.5^k– 5

Now, we will add and subtract by 3.7.5k- 7.5,

or, 2.7.7^k + 3.5.5^k – 5 + 3.7.5^k – 7.5 – (3.7.5^k – 7.5)

or, 7(2.7^k + 3.5^k – 5) + 3.5.5^k – 5 – (3.7.5^k – 7.5)

Now from equation (1) we have

or, 7(24q) + 3.5.5k- 5 – 3.7.5k+ 7.5

or, 7(24q) – 2.3.5k- 5 + 35

or, 7(24q) – 2.3.5k+ 30

or, 7(24q) – 6(5k- 5)

Now, we know that (5k – 5) is a multiple of 4. 

So we can write (4p) where (p) is from natural number.

or, 7(24q) – 6(4p)

or, 24(7q – p)

So, P (n) is true for n = k + 1

i.e., P (n) is true for all n ∈ N

Hence, by principle of Mathematical Induction (PMI), P (n) is true for all n ∈ N.

问题 27。对于 All N ∈ N，11 ⁿ⁺² + 12 ²ⁿ⁺¹可以被 133 整除。

解决方案：

Let P (n) = 11ⁿ⁺²+ 12²ⁿ⁺¹ is Divisible by 133

Step 1:

Now, let us check P(n) for n=1.

P (1) = 11¹⁺² + 12²⁺¹= 1331 + 1728 = 3059

So, P(1) is true as P (n) is divisible by 133.

Step 2:

Let us consider P (n) be the true for n = k,

P (k) = 11^k+2+ 12^2k+1 is Divisible by 133

or, 11^k+2+ 12^2k+1 = 133p  … (i)

Step 3:

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

i.e., 11^k+3+ 12^2k+3

= 11^k+2. 11 + 12^2k+1. 12² + 11. 12^2k+1 − 11. 12^2k+1

= 11 (11^k+2 + 12^2k+1) + 12^2k+1(144 − 11)

Now from equation (i) we have

= 11. 133p + 12^2k+1. 133               [From eq(i)]

= 133 (11p + 12^2k+1)

So, P (n) is true for n = k + 1

i.e., P (n) is true for all n ∈ N

Hence, by principle of Mathematical Induction (PMI), P (n) is true for all n ∈ N.

问题 28。1×1！ + 2×2！ + 3×3！ +…+ n×n！ = (n + 1)！ – 1 代表所有 n ∈ N

解决方案：

Let P (n) = 1×1! + 2×2! + 3×3! +…+ n×n

Step 1:

Now, let us check P(n) for n=1.

P (1) = 1 × 1! = (2)! – 1 = 1

So, P(1) is true.

Step 2:

Let us consider P (n) be the true for n = m,

P (k) = 1×1! + 2×2! + 3×3! +…+ m×m! = (m + 1)! – 1  … (i)

Step 3:

Now, we have to prove for P(m + 1) is true.

i.e., 1×1! + 2×2! + 3×3! +…+ m×m! + (m + 1)×(m + 1)!

So, = (m + 1)! – 1 + (m + 1) × (m + 1)!

= (m + 1)!(m + 1 + 1) – 1

= (m + 1)!(m + 2) – 1

= (m + 2)! – 1

So, P (n) is true for n = m + 1

i.e., P (n) is true for all n ∈ N

Hence, by principle of Mathematical Induction (PMI), P (n) is true for all n ∈ N.

问题 29. 证明对于所有 n ∈ N，n ³ – 7n + 3 可被 3 整除。

解决方案：

Let P(n) = n³ – 7n + 3 is divisible by 3 

Step 1:

Now, let us check P(n) for n=1.

P(1) = 1³ − (7 × 1) + 3 = 1 − 7 + 3 = −3

So, P(1) is true as P(n) is divisible by 3.

Step 2:

Let us consider P (n) be the true for n = k,

P (k) = k³ – 7k + 3 is divisible by 3

or, k³ – 7k + 3 = 3m  … (i)

Step 3:

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

i.e.,  = (k + 1)³ − 7 (k + 1) + 3

= k³ + 3k² + 3k + 1 − 7k − 7 + 3

= k³ + 3k² − 4k − 3

= k³ − 7k + 3 + 3k² + 3k − 6

Now from equation (i), we have

= 3m + 3 (k² + k + 2)

= 3 (m + k² + k + 2)

= 3t              (Here, t is any integer)

So, P (n) is true for n = k + 1

i.e., P (n) is true for all n ∈ N

Hence, by principle of Mathematical Induction (PMI), P (n) is true for all n ∈ N.

问题 30. 证明 1 + 2 + 2 ² + … + 2 ⁿ = 2 ⁿ⁺¹ – 1 对于所有 n ∈ N

解决方案：

Let P (n) = 1 + 2 + 2² + … + 2ⁿ = 2ⁿ⁺¹ – 1

Step 1:

Now, let us check P(n) for n = 1.

P (1) = 1 + 2 = 2¹⁺¹ – 1

or, 3 = 3

So, P(1) is true.

Step 2:

Let us consider P (n) be the true for n = k,

P (k) = 1 + 2 + 2² + … + 2^k = 2^k+1 – 1 … (i)

Step 3:

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

i.e., 1 + 2 + 2² + … + 2^k+1 = 2^k+2 – 1

So, 1 + 2 + 2² + … + 2^k + 2^k+1

Now from equation (i), we have

or, 2^k+1 – 1 + 2^k+1 

or, 2.2^k+1 – 1

or, 2^k+2 – 1 

So, P (n) is true for n = k + 1

i.e., P (n) is true for all n ∈ N

Hence, by principle of Mathematical Induction (PMI), P (n) is true for all n ∈ N.

问题 31。 7 + 77 + 777 + ... + 777 。 . . . . . . . . . . n - 数字 7 = 7/81 (10 ⁿ⁺¹ - 9n - 10)

解决方案：

Let P (n) = 7 + 77 + 777 + … + 777 . . . . . . . . . . . n − Digits 7 = 7/81 (10 ⁿ⁺¹ − 9n − 10)

Step 1:

Now, let us check P(n) for n = 1.

P (1) = 7 = 7/81 (10² – 9 – 10)

or, 7 = 7/81 ( 100 – 19)

or, 7 = 7/81 (81)

or, 7 = 7

So, P(1) is true.

Step 2:

Let us consider P (n) be the true for n = k,

P (k) =  7 + 77 + 777 + … + 777 . . . . . . . . . . . k− Digits7 = 7/81 (10 ^k+1 − 9k − 10)  … (i)

Step 3:

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

i.e., 7 + 77 + 777 + … + 777 . . . . . . . . . . . (k + 1) − Digits7

So, it is basically G.P. (geometric progression) with k + 1 terms.

So, sum P(k + 1)

or, 7/9[9 + 99 + 999 + ……………..+ (k + 1) term]

or, 7/9 [10 – 1 + 100 – 1 +………………..+ (k + 1)term]

or, 7/9 [10 + 100 + 1000 +…………+ (k + 1)term – (1 + 1 + 1 +….+ (k + 1) times)]

or, 7/81 [10^k+2 – 9(k+1) – 10]

So, P (n) is true for n = k + 1

i.e., P (n) is true for all n ∈ N

Hence, by principle of Mathematical Induction (PMI), P (n) is true for all n ∈ N.

问题 32. 证明 (n ⁷ /7) + (n ⁵ /5) + (n ³ /3) + (n ² /2) – (37/210)n 是所有 n ∈ N 的正整数。

解决方案：

Let P(n) be (n⁷/7) + (n⁵/5) + (n³/3) + (n²/2) – (37/210)n

Step 1:

Now, let us check P(n) for n = 1.

So, P(1) = 1/7+1/5+1/3+1/2-37/210 =1

So, P(1) is true.

Step 2:

Let us consider P (n) be the true for n = m,

So, let P(m) be (m⁷/7) + (m⁵/5) + (m³/3) + (m²/2) – (37/210)m

Now, (m⁷/7) + (m⁵/5) + (m³/3) + (m²/2) – (37/210)m = λ, where λ ∈ N is a positive integer.

Step 3:

Now, we have to prove for P(m+1) is true.

P(m + 1) = ((m + 1)⁷/7) + ((m + 1)⁵/5) + ((m + 1)³/3) + ((m + 1)²/2) – (37/210)(m + 1)

= 1/7(m⁷ + 7m⁶ + 21m⁵ + 35m⁴ + 35m³ + 21m² + 7m + 1) +

1/5(m⁵ + 5m⁴ + 10m³ + 10m² + 5m + 1) + 1/3(m³ + 3m² + 3m + 1) +

1/2(m² + 2m + 1) + 37/210(m + 1)

= (m⁷/7 + m⁵/5 + m³/3 + m²/2 − 37/210m) + m⁶ + 3m⁵ + 5m⁴ + 5m³ + 3m² +

 m + m⁴ + 2m³ + 2m² + m + m² + m + m + 1/7 + 1/5 + 1/3 + 1/2 – 37/210

= λ + 3m⁵ + 5m⁴ + 5m³ + 3m² + m + m⁴ + 2m³ + 2m² + m + m² + m + m + 1

As λ is positive, so it is a positive integer.

So, P (n) is true for n = m + 1

i.e., P (n) is true for all n ∈ N

Hence, by principle of Mathematical Induction (PMI), P (n) is true for all n ∈ N.