第 11 课 RD Sharma 解决方案 - 第 12 章数学归纳 - 练习 12.2 |设置 3

问题 33. 证明 n ¹¹ /11 + n ⁵ /5 + n ³ /3 – 62/165n 对于所有 n ∈ N 都是正确的。

解决方案：

Let, P(n) = n¹¹/11 + n⁵/5 + n³/3 – 62/165n

Step 1:

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

So, P(1) = 1/11 + 1/5 + 1/3 – 62/165 = 1

So, P(1) is true.

Step 2:

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

Let, P(m) = m¹¹/11 + m⁵/5 + m³/3 – 62/165m

So, m¹¹/11 + m⁵/5 + m³/3 – 62/165m = 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)¹¹/11 + (m + 1)⁵/5 + (m + 1)³/3 – 62/165(m + 1)

= 1/11(m¹¹ + 11m¹⁰ + 55m⁹ + 165m⁸ + 330m⁷ + 462m⁶ + 462m⁵ + 330m⁴ + 

165m³ + 55m² + 11m + 1) + 1/5(m⁵ + 5m⁴ + 10m³ + 10m² + 5m + 1) + 

1/3(m³ + 3m² + 3m + 1) + 62/165(m + 1)

= λ + m⁶ + 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.

问题 34. 证明 (1/2)tan(x/2) + (1/4)tan(x/4) +…..+ (1/2 ⁿ )tan(x/2 ⁿ ) = (1/ 2 ⁿ )cot(x/2 ⁿ ) – 所有 n ∈ N 和 0< x < π/2 的 cotx。

解决方案：

Let, P(n) = (1/2)tan(x/2) + (1/4)tan(x/4) +…..+ (1/2ⁿ)tan(x/2ⁿ) = (1/2ⁿ)cot(x/2ⁿ) – cotx, for all n ∈ N and 0< x < π/2.

Step 1:

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

LHS = 1/2 tan(x/2)

RHS = (1/2)cot(x/2) – cot(x) = 1/(2tan(x/2)) – 1 tan(x)

= 1/(2tan(x/2)) – 1/(2tan(x/2))/(1 – tan²(x/2))

= 1/(2tan(x/2)) – (1 – tan²(x/2))/(2tan(x/2))

=(1/2)tan(x/2)

So, P(1) is true.

Step 2:

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

P(m) = (1/2)tan(x/2) + (1/4)tan(x/4) +…..+ (1/2^m)tan(x/2^m) = (1/2^m)cot(x/2^m) – cotx

Step 3:

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

P(m + 1) = (1/2)tan(x/2) + (1/4)tan(x/4) +…..+  (1/2^m)tan(x/2^m) + (1/2^{(m + 1)})tan(x/2^{(m + 1)}) = (1/2^{(m + 1)})cot(x/2^{(m + 1)}) – cotx

So, = (1/2^m)cot(x/2^m) – cotx + (1/2^(m+1))cot(x/2^(m+1))

= (1/2^m)cot (x/2^m) + (1/2^(m+1))tan(x/2^(m+1)) – cotx

= 1/(2^mtan(2x/2^(m+1)) + (1/2^(m+1))tan(x/2^(m+1)) – cotx

= [(1 – tan²(x/2^(m+1)))/2^(m+1).tan(x/2^(m+1))] + (1/2^(m+1))tan(x/2^(m+1)) – cotx

= (1/2^(m+1))cot(x/2^(m+1)) – cotx

Now,

(1/2)tan(x/2) + (1/4)tan(x/4)+…..+ (1/2^m)tan(x/2^m) + (1/2^(m+1))tan(x/2^(m+1)) = (1/2^(m+1))cot(x/2^(m+1)) – cotx

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 35. 证明 (1 – 1/2 ² )(1 – 1/3 ² )(1 – 1/4 ² )……(1 – 1/n ² ) = (n + 1)/2n 对于所有 n ∈ N。

解决方案：

Let, P(n) =  (1 – 1/2²)(1 – 1/3²)(1 – 1/4²)……(1 – 1/n²) = (n + 1)/2n

Step 1:

Now, let us check P(n) for n = 2.

P(2) = 1 – 1/2² = (2 + 1)/2.2

or, 3/4 = 3/4

So, P(2) is true.

Step 2:

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

P(k) =  (1 – 1/2²)(1 – 1/3²)(1 – 1/4²)……(1 – 1/k²) = (k + 1)/2k

Step 3:

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

P(k + 1) = (1 – 1/2²)(1 – 1/3²)(1 – 1/4²)……(1 – 1/(k + 1)²) = (k + 2)/2k

Now, (1 – 1/2²)(1 – 1/3²)(1 – 1/4²)……(1 – 1/k²)(1 – 1/(k + 1)²)

Now, from step 2, we get

So, (1 – 1/(k + 1)²)((k + 1)/2k)

or, ((k + 1)/2k)((k² + 1 + 2k – 1)/(k + 1)²) 

or, k(k + 2)/2k(k + 1)

or, (k + 2)/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 36. 证明 (2n)!/2 ²ⁿ (n!) ² ≤ 1/√(3n + 1) 对所有 n ∈ N 成立。

解决方案：

Let, P(n) = (2n)!/2²ⁿ(n!)² ≤ 1/√(3n + 1) 

Step 1:

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

P(1) = 1/2 ≤ 1/√3 + 1 = 1/2

So, P(1) is true.

Step 2:

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

P(m) = (2m)!/2^2m(m!)² ≤ 1/√(3m + 1) 

Step 3:

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

P(m + 1) = (2m + 2)!/2^2m+2(m!)² ≤ 1/√(3m + 4)  

Now,

P(m + 1) = ((2m + 1)(2m + 1)(2m)!)/(2m².2²(m + 1)²(m!)²)

or, (2m + 2)!/2{2m + 2}((m + 1)!)² = ((2m)!/2^2m) × (2m + 1)(m + 2)/2²(m + 1)²

or, (2m + 2)!/2{2m + 2}((m + 1)!)² ≤ (2m + 1)/2(m + 1)√(3m + 1)

or, (2m + 2)!/2{2m + 2}((m + 1)!)² ≤ √((2m + 1)²/4(m + 1)²(3m + 1))

(2m + 2)!/2{2m + 2}((m + 1)!)² ≤ √((2m + 1)²/4(m + 1)²(3m + 1))

or, (2m + 2)!/2{2m + 2}((m + 1)!)² ≤ √(12m³ + 28m² + 19m + 4)/(12m³ + 28m² + 20m + 4)(3m + 4)

now, (12m³ + 28m² + 19m + 4)/(12m³ + 28m² + 20m + 4) < 1

so, (2m + 2)!/2{2m + 2}((m + 1)!)² ≤ 1/√(3m + 4)

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 37. 证明 1 + 1/4 + 1/9 + 1/16 + .....+ 1/n ² < 2 – 1/n 对于所有 n > 2，n ∈ N。

解决方案：

Let P(n) = 1 + 1/4 + 1/9 + 1/16 + …..+ 1/n² < 2 – 1/n for all n > 2, n ∈ N

Step 1:

Now, let us check P(n) for n = 2.

P(2) = 1/2² < 2 – 1/2

So, P(1) is true.

Step 2:

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

P(m) = 1 + 1/4 + 1/9 + 1/16 +…..+ 1/m² < 2 – 1/m

Step 3:

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

From step 2 we have, 1 + 1/4 + 1/9 + 1/16 +…..+ 1/m² < 2 – 1/m 

Now, adding 1/(m + 1)² to both the sides, we get

= 1 + 1/4 + 1/9 + 1/16 +…..+ 1/m² + 1/(m + 1)² < 2 – 1/m + 1/(m + 1)²

or, (m + 1)² > m + 1

or, 1/(m + 1)² < 1/(m + 1)

or, 1/m – 1/(m + 1)² < 1/(m + 1)

So, P(m + 1) < 2 – 1/(m + 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 38. 证明 x ^2n-1 + y ^2n-1可以被 x + y 整除。

解决方案：

Let, P(n) be   x^2n-1 + y^2n-1

Step 1:

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

P(1) = x + y  

So, P(1) is divisible by x + y.

So, P(1) is true.

Step 2:

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

P(m) = x^2m-1 + y^2m-1= λ(x + y)         ……….. (i)

Step 3:

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

i.e. ,  P(m + 1) = x^2m+1 + y^2m+1

= x^2m+1 + y^2m+1 – x^2m-1. y²  +  x^2m-1 . y²

= x^2m-1(x² – y²) + y²(x^2m-1+y^2m-1)

= (x + y)(x^2m-1 (x – y) + λy²)

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 39. 证明 sinx + sin3x + …+ sin(2n – 1)x = sin ² nx/sinx 对于所有 n ∈ N 都是正确的。

解决方案：

Let, P(n) =  sinx + sin3x + …+ sin(2n – 1)x = sin²nx/sinx

Step 1:

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

P(1) = sin x = sin² x / sin x = sin x

So, P(1) is true.

Step 2:

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

P(m) =  sinx + sin3x + …+ sin(2m – 1)x = sin²mx/sinx

Step 3:

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

i.e., 

P(m + 1) = sinx + sin3x +…+ sin(2m – 1)x + sin(2m + 1)x = (sin²mx/sinx) + sin(2m+1)x

now,

P(m + 1) = {(sin²mx + sinx[sin(mx) + cos(m + 1)x + sin(m + 1)x + cos(mx)])}/sinx

= {(sin²mx + 2sinx.cosx.cos(mx) – sin²x.sin²mx + cos²mx.sin²x)}/sinx

= {(sin²mx(1 – sin²x) + 2sinx.cosx.cos(mx) + cos²mx.sin²x)}/sinx

= {(sin²mx.cos²x + 2sinx.cosx.cos(mx) + cos²mx.sin²x)}/sinx

= {(sin(mx).cosx + cos(mx).sinx)²}/sinx

= {(sin(m + 1)x)²}/sinx  

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 40. 证明$cosα + cos(α +β) +cos(α +2β)+....+cos(α +(n-1)β)=\frac{(cos[α +(\frac{n-1}{2})β] sin(\frac{nβ}{2}))}{sin(\frac{β}{2})}$对所有 n ∈ N 都成立。

解决方案：

Let, P(n) = $cosα + cos(α +β) +cos(α +2β)+....+cos(α +(n-1)β)=\frac{(cos[α +(\frac{n-1}{2})β] sin(\frac{nβ}{2}))}{sin(\frac{β}{2})}$

Step 1:

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

L.H.S = cos [α + (1 – 1)β] = cos α

R.H.S =$\frac{(cos[α +(\frac{1-1}{2})β] sin(\frac{β}{2}))}{sin(\frac{β}{2})} = cosα$

So, P(1) is true as LHS and RHS are equal.

Step 2:

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

P(k) = $cosα + cos(α +β) +cos(α +2β)+....+cos(α +(k-1)β)=\frac{(cos[α +(\frac{k-1}{2})β] sin(\frac{kβ}{2}))}{sin(\frac{β}{2})}$

Step 3:

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

So, adding cos(α + kβ) both sides of P(k), we get

$P(k+1)= cosα + cos(α +β) +cos(α +2β)+....+cos(α +(k-1)β) +cos(α +kβ)=\frac{(cos[α +(\frac{k-1}{2})β] sin(\frac{kβ}{2}))}{sin(\frac{β}{2})}+cos(α +kβ)$

$=\frac{(cos[α +(\frac{k-1}{2})β] sin(\frac{kβ}{2})+cos(α +kβ)sin(\frac{β}{2})}{sin(\frac{β}{2})}$

$=\frac{(-sin(α-\frac{β}{2}) + sin(α+kβ+\frac{β}{2}))}{2sin(\frac{β}{2})}$

$=\frac{2cos(\frac{(2α +kβ)}{2})sin(\frac{(kβ+β)}{2})}{2sin(\frac{β}{2})}$

$=\frac{cos(α +\frac{kβ}{2})sin(\frac{(k+1)β}{2})}{sin(\frac{β}{2})}$

Now,

$RHS= \frac{(cos[α +(\frac{kβ}{2})] sin(\frac{(k+1)β}{2}))}{sin(\frac{β}{2})}$

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 41. 证明 1/(n+1) + 1/(n+2) +…..+ 1/2n > 13/24 对于所有自然数，n > 1。

解决方案：-

Let,P(n) = 1/(n+1) + 1/(n+2) +…..+ 1/2n > 13/24

Step 1:

Now, let us check P(n) for n = 2.

P(2) = 1/(2 + 1) + 1/(2 + 2) = 1/3 + 1/4 = 7/12 > 13/24

So, P(2) is true.

Step 2:

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

P(k) = 1/(k + 1) + 1/(k + 2) +…..+ 1/2k > 13/24

Step 3:

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

So, P(k + 1) = 1/(k + 2) + 1/(k + 3) +…..+ 1/2k + 1/2(k + 1)

Here, as LHS = RHS

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 42. 给定 a ₁ = 1/2(a ₀ + A/a ₀ )，a ₂ = 1/2(a ₁ + A/a ₁ ) 和 a _n+1 = 1/2(a _n + A/ a _n ), a, A > 0

证明： $\frac{(a_n - √A)}{(a_n + √A)} = (\frac{ a_1 - √A}{a_1 + √A})^{2^{(n-1)}}$

解决方案：

Let, P(n) = $\frac{(a_n - √A)}{(a_n + √A)} = (\frac{ a_1 - √A}{a_1 + √A})^{2^{(n-1)}}$

Step 1:

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

LHS = (a₁ – √A) / (a₁ + √A) 

RHS = $(\frac{ a_1 - √A}{a_1 + √A})^{2^{(1-1)}}$

= (a₁ – √A) / (a₁ + √A)

So, P(1) is true.

Step 2:

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

So, P(k) = $\frac{(a_k - √A)}{(a_k + √A)} = (\frac{ a_1 - √A}{a_1 + √A})^{2^{(k-1)}}$

Step 3:

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

So, P(k + 1) = LHS = (a_k+1 – √A) / (a_k+1 + √A)

=  (1/2(a_k + A/a_k) – √A) /  (1/2(a_k + A/a_k) + √A) 

= (1/2(a_k²+ A – 2a_k√A)/a_k ) / (1/2(a_k²+ A + 2a_k√A)/a_k )

=  (a_k+1 – √A)² / (a_k+1 + √A)²

= $[ [\frac{a_1 - √A}{a_1 + √A}]^{2^{k-1}}]^2$ 

= $[\frac{a_1 - √A}{a_1 + √A}]^{2^k}$

here, as LHS = RHS

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 43. 设 P(n) 为陈述：2 ⁿ ≥ 3n。如果 P(r) 为真，则证明 P(r + 1) 为真，你是否得出结论 P(n) 对所有 n ∈ N 都为真？

解决方案：

Let P(n) = 2ⁿ ≥ 3n

Step 1:

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

L.H.S = 2

R.H.S = 3

As L.H.S < R.H.S

So, P(1) is true.

Step 2:

Let us consider P (n) be the true for n = r, So, P(r) is 2^r ≥ 3r

Step 3:

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

So, P(k + 1) = 2^r+1 = 2.2^r

For, x > 3, 2x > x + 3

So, 2.2^r > 2^r+3 for r > 1

or,2^r+1 > 2^r+3 for r > 1

or, 2^r+1 > 3r +3 for r > 1

or, 2^r+1 > 3(r + 1) for r >1

So, P (n) is true for n = r + 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 44. 用数学归纳法证明级数 1 ² + 2 × 2 ² + 3 ² + 2 × 4 ² + 5 ² + 2 × 6 ² + 7 ² +… 的 n 项之和 Sn 是由

$S_n= \begin{cases} \frac{n(n+1)^2}{2}, if \ n \ is \ even, \\ \frac{n^2(n+1)}{2}, if \ n \ is \ odd \end{cases}$

解决方案：

Let, P(n) = S_n = 1² + 2 × 2² + 3² + 2 × 4² + 5² + 2 × 6² + 7² +… = $\begin{cases} \frac{n(n+1)^2}{2}, if \ n \ is \ even, \\ \frac{n^2(n+1)}{2}, if \ n \ is \ odd \end{cases}$

Step 1:

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

LHS = 1 = RHS

So, P(1) is true.

Step 2:

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

P(k) = 1² + 2.2² + 3² + 2.4² + 5² = $\begin{cases} \frac{k(k+1)^2}{2}, if \ k \ is \ even, \\ \frac{k^2(k+1)}{2}, if \ k \ is \ odd \end{cases}$

Step 3:

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

Case1: When k is odd, then (k + 1) is even

P(k + 1) = LHS = 1² + 2.2² + 3² + 2.4² + 5²…. k² + 2.(k + 1)²

= k²(k + 1)/2 + 2.(k + 1)²

= {(k²(k + 1) + 4(k + 1)²)}/2

= (k + 1)(k + 2)²/2

now, RHS = (k + 1)(k + 1 + 1)²/2

= (k + 1)(k + 2)²/2

So, it is true for n = k + 1, when k is odd.

Case 2: When k is even, then (k + 1) is odd

P(k + 1) = LHS = 1² + 2.2² + 3² + 2.4² + 5²….+ 2. k² + (k + 1)²

= k(k + 1)²/2 + (k + 1)²

= (k(k + 1)² + 2.(k + 1)²)/2

= (k + 1)²(k + 2)/2

RHS = (k + 1)²(k + 1 + 1)/2

= (k + 1)²(k + 2)/2

now, LHS = RHS.

So, it is true for n=k+1, when k is even.

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 45. 证明对于所有 n ∈ N，包含 n 个不同元素的集合的子集数为 2 ⁿ 。

解决方案：

Let, P(n): The number of subsets of a set containing n distinct elements = 2ⁿ, for all n ∈ N.

Step 1:

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

LHS = As, the subsets of the set containing only 1 element are:

Φ and the set itself

i.e. the number of subsets of a set containing only element=2

R.H.S = 2¹ = 2

now, LHS = RHS

So, P(1) is true.

Step 2:

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

So, P(k) is The number of subsets of a set containing k distinct elements = 2^k

Step 3:

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

P(k + 1) = Let A = {a₁, a₂, a₃, a₄,..…, a_k, b} so that A has (k + 1) elements.

Now, the subset t of A can be divided into two collections such that 

First contains subsets of A which don’t have b in them and

The second contains subsets of A which do have b in them.

So, First collection: {}, {a₁}, {a₁, a₂}, {a₁, a₂, a₃},…,{a₁, a₂, a₃, a₄,…, a_k} and

and the second collection: {b}, {a₁, b}, {a₁, a₂, b}, {a₁, a₂, a₃, b},…,{a₁, a₂, a₃, a₄,…,a_k, b}

It can be clearly seen that:

The number of subsets of A in first collection

= The number of subsets of set with k elements i.e. {a₁, a₂, a₃, a₄,…, a_k} = 2k

Also, it follows that the second collection must have

the same number of the subsets as that of the first = 2^k

So the total number of subsets of A = 2^k + 2^k = 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 46. 序列 a ₁ , a ₂ , a ₃ ,….. 定义为对于所有自然数 k ≥ 2 的 a ₁ = 3 和 a _k = 7a _k-1 。证明 a _n = 3.7 ^{n- 1}对于所有 n ∈ N。

解决方案：

Let P(n) be a_n = 3.7^n-1 for all n ∈ N

Step 1: 

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

so, a₁ = 3.7^1-1 = 3

So, P(1) is true.

Step 2:

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

P(k) = a_k = 3.7^k-1

Step 3:

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

P(k + 1) = a_k+1 = 7.a_k

= 7.3.7^k-1

= 3.7^k-1+1

= 3.7^(k+1)-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 47. 序列 x ₁ , x ₂ , x ₃ ,….. 定义为 x ₁ = 2 和 x _k = x _k-1 /k 对于所有自然数 k, k ≥ 2。证明 x _n = 2/n！对于所有 n ∈ N。

解决方案：

Let, P(n) be x_n = 2/n!  for all n ∈ N.

Step 1:

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

P(1) = x₁ = 2/1! = 2

So, P(1) is true.

Step 2:

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

P(k) = x_k = 2/k! 

Step 3:

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

so,  x_k+1 = 2/(k + 1)! 

or, 2/(k + 1).k!

or, 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 48. 一个序列 x ₀ , x ₁ , x ₂ , x ₃ , ……。是通过让 x ₀ = 5 和 x _k = 4 + x _{k -1}对于所有自然数 k 来定义的。使用数学归纳法证明对于所有 n ∈ N，x _n = 5 + 4 _n 。

解决方案：-

Let, P(n) = x_n = 5 + 4n for all n ∈ N

Step 1:

Now, let us check P(n) for n = 0.

P(0) = 5 + 4(0) = 5

So, P(0) is true.

Step 2:

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

P(k) = x_k = 5 + 4k  

Step 3:

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

so, P(k + 1) = x_k+1 = 4 + x_{k+1 -1}

= 4 + x_k

= 4 + 5 + 4k

= 5 + 4(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 49. 利用数学归纳原理证明

√n < 1/√1 + 1/√2+ 1/√3 +…..+1/√n 对于所有自然数 n ≥2。

解决方案：

Let, P(n) = √n < 1/√1 + 1/√2+ 1/√3 +…..+1/√n for all n ≥ 2.

Step 1:

Now, let us check P(n) for n=2.

P(2) = √2 < 1 + 1/√2

or, 1.41 < 1 + 0.707 = 1.707

So, P(2) is true.

Step 2:

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

P(k) = √k < 1/√1 + 1/√2+ 1/√3 +…..+1/√k

Step 3:

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

Now, LHS = √(k + 1)

Now, RHS = 1/√1 + 1/√2+ 1/√3 +…..+1/√k + 1/√(k + 1)

or, k/√{(k + 1)} < √k

or, k + 1/√{(k + 1)} – 1/√(k + 1) < √k

or, √(k + 1) – 1/√(k + 1)< √k

or, √(k + 1) < √k + 1/√(k + 1)

so, LHS < RHS.

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

i.e., P (n) is true for all n ≥ 2

Hence, by principle of Mathematical Induction (PMI), P (n) is true for all n ≥ 2.

问题 50. 代数的分配定律指出，对于所有实数 a ₁和 a ₂ ，我们有 c (a ₁ + a ₂ ) = ca ₁ + ca ₂ 。用这个定律和数学归纳法证明，对于所有自然数，n ≥ 2，如果 c, a ₁ , a ₂ ,...,a _n是任意实数，则 c (a ₁ +a ₂ +...+a _n ) = ca ₁ +ca ₂ +…+ca _n 。

解决方案：

Let, P(n) = c (a₁+a₂+…+a_n) = ca₁+ca₂+…+ca_n, for all natural numbers, n ≥ 2. 

Step 1: 

Now, let us check P(n) for n=2.

LHS = c(a₁ + a₂)

RHS = c a₁ + ca₂

So, P(2) is true.

Step 2:

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

P(k) =  c(a₁+a₂+…+a_k) = ca₁+ca₂+…+ca_k

Step 3:

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

LHS = c(a₁+a₂+…+a_k + a_k+1)

= c[(a₁+a₂+…+a_k) + a_k+1]

= c(a₁+a₂+…+a_k) + ca_k+1

= ca₁+ca₂+…+ca_k + ca_k+1

= RHS

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

i.e., P (n) is true for all n ≥ 2

Hence, by principle of Mathematical Induction (PMI), P (n) is true for all n ≥ 2.