第 11 课 RD Sharma 解决方案 - 第 12 章数学归纳 - 练习 12.2 |设置 1

用数学归纳原理证明如下：

问题1. 1 + 2 + 3 + … + n = n (n +1)/2 即前n个自然数之和为n (n + 1)/2。

解决方案：

Let, P (n) = 1 + 2 + 3 + ….. + n = n (n +1)/2.

Step 1:

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

So, P (1) = 1 (1 + 1)/2 = 1.

So, P(1) is true.

Step 2:

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

Then, 1 + 2 + 3 + …. + k = k (k + 1)/2           … (i)

Step 3:

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

(1 + 2 + 3 + … + k) + (k + 1) = k (k + 1)/2 + (k + 1)

= (k + 1) (k/2 + 1)

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

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

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 2。1 ² + 2 ² + 3 ² + … + n ² = [n (n + 1) (2n + 1)]/6

解决方案：

Let, P (n) = 1² + 2² + 3² + … + n² = [n (n + 1) (2n + 1)]/6

Step 1:

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

So, P (1) = [1 (1 + 1) (2 + 1)]/6

or, 1 = 1.

So, P(1) is true.

Step 2:

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

Then, 1² + 2² + 3² + … + k² = [k (k + 1) (2k + 1)]/6

Step 3:

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

P (k) = 1² + 2² + 3² + …….. + k² + (k + 1)² = [k + 1 (k + 2) (2k + 3)] /6

= 1² + 2² + 3² + ……. + k² + (k + 1)²

= [k + 1 (k + 2) (2k + 3)] /6 + (k + 1)²

= (k +1) [(2k² + k)/6 + (k + 1)/1]

= (k +1) [2k² + k + 6k + 6]/6

= (k +1) [2k² + 7k + 6]/6

= (k +1) [2k² + 4k + 3k + 6]/6

= (k +1) [2k(k + 2) + 3(k + 2)]/6

= [(k +1) (2k + 3) (k + 2)] / 6

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 3。1 + 3 + 3 ² + … + 3n – 1 = (3n – 1)/2

解决方案：

Let P (n) = 1 + 3 + 3² + ……. + 3n – 1 = (3n – 1)/2

Step 1:

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

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

So, P(1) is true.

Step 2:

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

P (k) = 1 + 3 + 3² + ……. + 3k – 1 = (3k – 1)/2       … (i)

Step 3:

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

Then, 

{1 + 3 + 3² + ……. + 3k – 1} + 3k + 1 – 1

Using equation (i), we get

= (3k – 1)/2 + 3k 

= (3k – 1 + 2 × 3k)/2

= (3 × 3 k – 1)/2

= (3^{k + 1} – 1)/2

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 4. 1/1.2 + 1/2.3 + 1/3.4 + … + 1/n(n + 1) = n/(n + 1)

解决方案：

Let P (n) = 1/1.2 + 1/2.3 + 1/3.4 + … + 1/n(n + 1) = n/(n + 1)

Step 1:

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

P (1) = 1/1.2 = 1/(1 + 1)

1/2 = 1/2

So, P(1) is true.

Step 2:

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

P (k) = 1 /1.2 + 1/2.3 + 1/3.4 + … + 1/k(k + 1) + k/(k + 1) (k + 2) = (k + 1)/(k + 2)   … (i)

Step 3:

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

Then,

1/1.2 + 1/2.3 + 1/3.4 + … + 1/k(k + 1) + k/(k + 1) (k + 2)

= 1/(k + 1)/(k + 2) + k/(k + 1)

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

= 1/(k + 1) [k² + 2k + 1]/(k + 2)

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

= (k + 1) / (k + 2)

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 5. 1 + 3 + 5 + … + (2n – 1) = n ²即前 n 个奇数自然数之和为 n ² 。

解决方案：

Let P (n) = 1 + 3 + 5 + … + (2n – 1) = n²

Step 1:

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

P (1) = 1 = 1² = 1

So, P(1) is true.

Step 2:

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

P (k) = 1 + 3 + 5 + … + (2k – 1) = k²  … (i)

Step 3:

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

Then, 1 + 3 + 5 + … + (2k – 1) + 2(k + 1) – 1

= k² + (2k + 1)

= k² + 2k + 1

= (k + 1)²

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 6。1/2.5 + 1/5.8 + 1/8.11 + … + 1/(3n – 1) (3n + 2) = n/(6n + 4)

解决方案：

Let P (n) = 1/2.5 + 1/5.8 + 1/8.11 + … + 1/(3n – 1) (3n + 2) = n/(6n + 4)

Step 1:

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

P (1) = 1/2.5 = 1/(6.1 + 4)

or, 1/10 = 1/10

So, P(1) is true.

Step 2:

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

P (k) = 1/2.5 + 1/5.8 + 1/8.11 + … + 1/(3k – 1) (3k + 2) = k/(6k + 4) … (i)

Step 3:

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

Then, 1/2.5 + 1/5.8 + 1/8.11 + … + 1/(3k – 1)(3k + 2) + 1/(3k + 3 – 1)(3k + 3 + 2)

= k/(6k + 4) + 1/(3k + 2)(3k + 5)

= [k(3k + 5) + 2] / [2(3k + 2)(3k + 5)]

= (k + 1) / (6(k + 1) + 4)

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 7. 1/1.4 + 1/4.7 + 1/7.10 + … + 1/(3n – 2)(3n + 1) = n/3n + 1

解决方案：

Let P (n) = 1/1.4 + 1/4.7 + 1/7.10 + … + 1/(3n – 2)(3n + 1) = n/3n + 1

Step 1:

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

P (1) = 1/1.4 = 1/4

Or, 1/4 = 1/4

So, P(1) is true.

Step 2:

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

P (k) = 1/1.4 + 1/4.7 + 1/7.10 + … + 1/(3k – 2)(3k + 1) = k/3k + 1 … (i)

Step 3:

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

Then, [1/1.4 + 1/4.7 + 1/7.10 + … + 1/(3k – 2)(3k + 1)] + 1/(3k + 1)(3k + 4)

= k/(3k + 1) + 1/(3k + 1)(3k + 4)

= 1/(3k + 1) [k/1 + 1/(3k + 4)]

= 1/(3k + 1) [k(3k + 4) + 1]/(3k + 4)

= 1/(3k + 1) [3k² + 4k + 1]/ (3k + 4)

= 1/(3k + 1) [3k² + 3k + k + 1]/(3k + 4)

= [3k(k + 1) + (k + 1)] / [(3k + 4) (3k + 1)]

= [(3k + 1)(k + 1)] / [(3k + 4) (3k + 1)]

= (k + 1) / (3k + 4)

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 8. 1/3.5 + 1/5.7 + 1/7.9 + … + 1/(2n + 1)(2n + 3) = n/3(2n + 3)

解决方案：

Let P (n) = 1/3.5 + 1/5.7 + 1/7.9 + … + 1/(2n + 1)(2n + 3) = n/3(2n + 3)

Step 1:

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

P (1) = 1/3.5 = 1/3(2.1 + 3)

or, 1/15 = 1/15

So, P(1) is true.

Step 2:

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

P(k) = 1/3.5 + 1/5.7 + 1/7.9 + … + 1/(2k + 1)(2k + 3) = k/3(2k + 3) … (i)

Step 3:

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

Then, 1/3.5 + 1/5.7 + 1/7.9 + … + 1/(2k + 1)(2k + 3) + 1/[2(k + 1) + 1][2(k + 1) + 3]

1/3.5 + 1/5.7 + 1/7.9 + … + 1/(2k + 1)(2k + 3) + 1/(2k + 3)(2k + 5)

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

= k/3(2k + 3) + 1/(2k + 3)(2k + 5)

= [k(2k + 5) + 3] / [3(2k + 3)(2k + 5)]

= (k + 1) / [3(2(k + 1) + 3)]

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 9。1/3.7 + 1/7.11 + 1/11.15 + … + 1/(4n – 1)(4n + 3) = n/3(4n + 3)

解决方案：

Let P (n) = 1/3.7 + 1/7.11 + 1/11.15 + … + 1/(4n – 1)(4n + 3) = n/3(4n + 3)

Step 1:

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

P (1) = 1/(3.7) = 1/((4.1 – 1)(4 + 3))

Or, 1/21 = 1/21

So, P(1) is true.

Step 2:

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

P (k) = 1/3.7 + 1/7.11 + 1/11.15 + … + 1/(4k – 1)(4k + 3) = k/3(4k + 3) … (i)

Step 3:

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

Then, 1/3.7 + 1/7.11 + 1/11.15 + … + 1/(4k – 1)(4k + 3) + 1/(4k + 3)(4k + 7)

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

= k/(4k + 3) + 1/(4k + 3)(4k + 7)

= 1/(4k + 3) [k(4k + 7) + 3] / [3(4k + 7)]

= 1/(4k + 3) [4k² + 7k + 3]/ [3(4k + 7)]

= 1/(4k + 3) [4k² + 3k + 4k + 3] / [3(4k + 7)]

= 1/(4k + 3) [4k(k + 1) + 3(k + 1)]/ [3(4k + 7)]

= 1/(4k + 3) [(4k + 3)(k + 1)] / [3(4k + 7)]

= (k + 1) / [3(4k + 7)]

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 10. 1.2 + 2.2 ² + 3.2 ³ + … + n.2 ⁿ = (n – 1) 2 ⁿ⁺¹ + 2

解决方案：

Let P (n) = 1.2 + 2.2² + 3.2³ + … + n.2ⁿ = (n – 1) 2ⁿ⁺¹ + 2

Step 1:

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

P (1) = 1.2 = 0.2⁰ + 2

or, 2 = 2

So, P(1) is true.

Step 2:

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

P (k) =1.2 + 2.2² + 3.2³ + … + k.2^k = (k – 1) 2^k+1 + 2  … (i)

Step 3:

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

Then, [1.2 + 2.2² + 3.2³ + … + k.2^k] + (k + 1)2k + 1

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

= [(k – 1)2^k+1 + 2] + (k + 1)2^k+1 

= (k – 1)2^k+1 + 2 + (k + 1)2^k+1

= 2^k+1(k – 1 + k + 1) + 2

= 2^k+1 × 2k + 2

= k.2^k+2 + 2

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 11. 2 + 5 + 8 + 11 + ... + (3n – 1) = 1/2 n (3n + 1)

解决方案：

Let P (n) = 2 + 5 + 8 + 11 + … + (3n – 1) = 1/2 n (3n + 1)

Step 1:

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

P (1) = 2 = 1/2 × 1 × 4

or, 2 = 2

So, P(1) is true.

Step 2:

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

P (k) = 2 + 5 + 8 + 11 + … + (3k – 1) = 1/2 k (3k + 1)   … (i)

Step 3:

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

Then, 2 + 5 + 8 + 11 + … + (3k – 1) + (3k + 2)

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

= 1/2 × k (3k + 1) + (3k + 2)            (using equation (i))

= [3k² + k + 2 (3k + 2)] / 2

= [3k² + k + 6k + 2] / 2

= [3k² + 7k + 2] / 2

= [3k² + 4k + 3k + 2] / 2

= [3k (k + 1) + 4(k + 1)] / 2

= [(k + 1) (3k + 4)] /2

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 12。1.3 + 2.4 + 3.5 + ... + n。 (n + 2) = 1/6 n (n + 1) (2n + 7)

解决方案：

Let P (n) = 1.3 + 2.4 + 3.5 + … + n. (n + 2) = 1/6 n (n + 1) (2n + 7)

Step 1:

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

P (1) = 1.3 = 1/6 × 1 × 2 × 9

Or, 3 = 3

So, P(1) is true.

Step 2:

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

P (k) = 1.3 + 2.4 + 3.5 + … + k. (k + 2) = 1/6 k (k + 1) (2k + 7)    … (i)

Step 3:

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

Then, 1.3 + 2.4 + 3.5 + … + k. (k + 2) + (k + 1) (k + 3)

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

= 1/6 k (k + 1) (2k + 7) + (k + 1) (k + 3)            (using equation (i))

= (k + 1) [{k(2k + 7)/6} + {(k + 3)/1}]

= (k + 1) [(2k² + 7k + 6k + 18)] / 6

= (k + 1) [2k² + 13k + 18] / 6

= (k + 1) [2k² + 9k + 4k + 18] / 6

= (k + 1) [2k(k + 2) + 9(k + 2)] / 6

= (k + 1) [(2k + 9) (k + 2)] / 6

= 1/6 (k + 1) (k + 2) (2k + 9)

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 13。1.3 + 3.5 + 5.7 + … + (2n – 1) (2n + 1) = n(4n ² + 6n – 1)/3

解决方案：

Let P (n) = 1.3 + 3.5 + 5.7 + … + (2n – 1) (2n + 1) = n(4n² + 6n – 1)/3

Step 1:

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

P (1) = : (2.1 – 1) (2.1 + 1) = 1(4.1² + 6.1 – 1)/3

or, 1 × 3 = 1(4 + 6 – 1)/3

or, 3 = 3

So, P(1) is true.

Step 2:

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

P (k) = 1.3 + 3.5 + 5.7 + … + (2k – 1) (2k + 1) = k(4k² + 6k – 1)/3    … (i)

Step 3:

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

Then, 1.3 + 3.5 + 5.7 + … + (2k – 1) (2k + 1) + (2k + 1) (2k + 3)

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

= k(4k² + 6k – 1)/3 + (2k + 1) (2k + 3) (using equation (i))

= [k(4k² + 6k-1) + 3 (4k² + 6k + 2k + 3)] / 3

= [4k³ + 6k² – k + 12k² + 18k + 6k + 9] /3

= [4k³ + 18k² + 23k + 9] /3

= [4k³ + 4k² + 14k² + 14k + 9k + 9] /3

= [(k + 1) (4k² + 8k +4 + 6k + 6 – 1)] / 3

= [(k + 1) 4[(k + 1)² + 6(k + 1) -1]] /3

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 14. 1.2 + 2.3 + 3.4 + … + n(n + 1) = [n (n + 1) (n + 2)] / 3

解决方案：

Let P (n) = 1.2 + 2.3 + 3.4 + … + n(n + 1) = [n (n + 1) (n + 2)] / 3

Step 1:

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

P (1) = 1(1 + 1) = [1(1 + 1) (1 + 2)] /3

or, 2 = 2

So, P(1) is true.

Step 2:

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

P (k) = 1.2 + 2.3 + 3.4 + … + k(k+1) = [k (k + 1) (k + 2)] / 3     … (i)

Step 3:

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

Then, 1.2 + 2.3 + 3.4 + … + k(k + 1) + (k + 1) (k + 2)

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

= [k (k + 1) (k + 2)] / 3 + (k + 1) (k + 2) 

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

= [(k + 1) (k + 2) (k + 3)] /3

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 15. 1/2 + 1/4 + 1/8 + … + 1/2 ⁿ = 1 – 1/2 ⁿ

解决方案：

Let P (n) = 1/2 + 1/4 + 1/8 + … + 1/2ⁿ = 1 – 1/2ⁿ

Step 1:

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

P (1) = 1/2¹ = 1 – 1/2¹

or, 1/2 = 1/2

So, P(1) is true.

Step 2:

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

P (k) = : 1/2 + 1/4 + 1/8 + … + 1/2^k = 1 – 1/2^k    … (i)

Step 3:

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

Then, 1/2 + 1/4 + 1/8 + … + 1/2^k + 1/2^k+1

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

= 1 – 1/2^k + 1/2^k+1   (using eq(i))

= 1 – ((2 – 1)/2^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 16。1 ² + 3 ² + 5 ² + … + (2n – 1) ² = 1/3 n (4n ² – 1)

解决方案：

Let P (n) = 1² + 3² + 5² + … + (2n – 1)² = 1/3 n (4n² – 1)

Step 1:

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

P (1) = (2.1 – 1)² = 1/3 × 1 × (4 – 1)

or, 1 = 1

So, P(1) is true.

Step 2:

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

P (k) = 1² + 3² + 5² + … + (2k – 1)² = 1/3 k (4k² – 1)     … (i)

Step 3:

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

Then, 1² + 3² + 5² + … + (2k – 1)² + (2k + 1)²

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

Or, 1/3 k (4k² – 1) + (2k + 1)²                         (using eq(i))

= 1/3 k (2k + 1) (2k – 1) + (2k + 1)²

= (2k + 1) [{k(2k – 1)/3} + (2k + 1)]

= (2k + 1) [2k² – k + 3(2k + 1)] / 3

= (2k + 1) [2k² – k + 6k + 3] / 3

= [(2k + 1) 2k² + 5k + 3] /3

= [(2k + 1) (2k(k + 1)) + 3 (k + 1)] /3

= [(2k + 1) (2k + 3) (k + 1)] /3

= (k + 1)/3 [4k² + 6k + 2k + 3]

= (k + 1)/3 [4k² + 8k – 1]

= (k + 1)/3 [4(k + 1)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.