# Class 11 RD Sharma Solutions – Chapter 12 Mathematical Induction – Exercise 12.2 | Set 3

### Question 33. Prove that n^{11}/11 + n^{5}/5 + n^{3}/3 – 62/165n is true for all n ∈ N.

**Solution:**

Let, P(n) = n

^{11}/11 + n^{5}/5 + n^{3}/3 – 62/165nAttention reader! All those who say programming isn't for kids, just haven't met the right mentors yet. Join the

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free classeswhich will definitely help them in making a wise career choice in the future.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) = m

^{11}/11 + m^{5}/5 + m^{3}/3 – 62/165mSo, m

^{11}/11 + m^{5}/5 + m^{3}/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(m

^{11 }+ 11m^{10 }+ 55m^{9 }+ 165m^{8 }+ 330m^{7 }+ 462m^{6 }+ 462m^{5 }+ 330m^{4 }+165m

^{3 }+ 55m^{2 }+ 11m + 1) + 1/5(m^{5 }+ 5m^{4 }+ 10m^{3 }+ 10m^{2 }+ 5m + 1) +1/3(m

^{3 }+ 3m^{2 }+ 3m + 1) + 62/165(m + 1)= λ + m

^{6 }+ 3m^{5 }+ 5m^{4 }+ 5m^{3 }+ 3m^{2 }+ m + m^{4 }+ 2m^{3 }+ 2m^{2 }+ m + m^{2 }+ m + m + 1As 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.

### Question 34. Prove that (1/2)tan(x/2) + (1/4)tan(x/4) +…..+ (1/2^{n})tan(x/2^{n}) = (1/2^{n})cot(x/2^{n}) – cotx for all n ∈ N and 0< x < π/2.

**Solution:**

Let, P(n) = (1/2)tan(x/2) + (1/4)tan(x/4) +…..+ (1/2

^{n})tan(x/2^{n}) = (1/2^{n})cot(x/2^{n}) – 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 – tan

^{2}(x/2))= 1/(2tan(x/2)) – (1 – tan

^{2}(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}) – cotxStep 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)}) – cotxSo, = (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

^{m}tan(2x/2^{(m+1)}) + (1/2^{(m+1)})tan(x/2^{(m+1)}) – cotx= [(1 – tan

^{2}(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)}) – cotxNow,

(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)}) – cotxSo, 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.

### Question 35. Prove that (1 – 1/2^{2})(1 – 1/3^{2})(1 – 1/4^{2})……(1 – 1/n^{2}) = (n + 1)/2n is true for all n ∈ N.

**Solution:**

Let, P(n) = (1 – 1/2

^{2})(1 – 1/3^{2})(1 – 1/4^{2})……(1 – 1/n^{2}) = (n + 1)/2nStep 1:

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

P(2) = 1 – 1/2

^{2}= (2 + 1)/2.2or, 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 – 1/2

^{2})(1 – 1/3^{2})(1 – 1/4^{2})……(1 – 1/k^{2}) = (k + 1)/2kStep 3:

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

P(k + 1) = (1 – 1/2

^{2})(1 – 1/3^{2})(1 – 1/4^{2})……(1 – 1/(k + 1)^{2}) = (k + 2)/2kNow, (1 – 1/2

^{2})(1 – 1/3^{2})(1 – 1/4^{2})……(1 – 1/k^{2})(1 – 1/(k + 1)^{2})Now, from step 2, we get

So, (1 – 1/(k + 1)

^{2})((k + 1)/2k)or, ((k + 1)/2k)((k

^{2 }+ 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.

### Question 36. Prove that (2n)!/2^{2n}(n!)^{2} ≤ 1/√(3n + 1) is true for all n ∈ N.

**Solution:**

Let, P(n) = (2n)!/2

^{2n}(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)!)/(2m

^{2}.2^{2}(m + 1)^{2}(m!)^{2})or, (2m + 2)!/2{2m + 2}((m + 1)!)

^{2}= ((2m)!/2^{2m}) × (2m + 1)(m + 2)/2^{2}(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}≤ √(12m^{3 }+ 28m^{2 }+ 19m + 4)/(12m^{3 }+ 28m^{2 }+ 20m + 4)(3m + 4)now, (12m

^{3 }+ 28m^{2 }+ 19m + 4)/(12m^{3 }+ 28m^{2 }+ 20m + 4) < 1so, (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.

### Question 37. Prove that 1 + 1/4 + 1/9 + 1/16 + …..+ 1/n^{2} < 2 – 1/n for all n > 2, n ∈ N.

**Solution:**

Let P(n) = 1 + 1/4 + 1/9 + 1/16 + …..+ 1/n

^{2}< 2 – 1/n for all n > 2, n ∈ NStep 1:

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

P(2) = 1/2

^{2}< 2 – 1/2So, 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/m

^{2}< 2 – 1/mStep 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/m

^{2}< 2 – 1/mNow, adding 1/(m + 1)

^{2}to both the sides, we get= 1 + 1/4 + 1/9 + 1/16 +…..+ 1/m

^{2 }+ 1/(m + 1)^{2}< 2 – 1/m + 1/(m + 1)^{2}or, (m + 1)

^{2}> m + 1or, 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.

### Question 38. Prove that x^{2n-1} + y^{2n-1} is divisible by x + y.

**Solution:**

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}. y^{2 }+ x^{2m-1 }. y^{2}= x

^{2m-1}(x^{2 }– y^{2}) + y^{2}(x^{2m-1}+y^{2m-1})= (x + y)(x

^{2m-1 }(x – y) + λy^{2})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.

### Question 39. Prove that sinx + sin3x + …+ sin(2n – 1)x = sin^{2}nx/sinx is true for all n ∈ N.

**Solution:**

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

^{2}nx/sinxStep 1:

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

P(1) = sin x = sin

^{2}x / sin x = sin xSo, 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 = sin

^{2}mx/sinxStep 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 = (sin

^{2}mx/sinx) + sin(2m+1)xnow,

P(m + 1) = {(sin

^{2}mx + sinx[sin(mx) + cos(m + 1)x + sin(m + 1)x + cos(mx)])}/sinx= {(sin

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

^{2}mx(1 – sin^{2}x) + 2sinx.cosx.cos(mx) + cos^{2}mx.sin^{2}x)}/sinx= {(sin

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

^{2}}/sinx= {(sin(m + 1)x)

^{2}}/sinxSo, 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.

### Question 40. Prove that is true for all n ∈ N.

**Solution:**

Let, P(n) =

Step 1:

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

L.H.S = cos [α + (1 – 1)β] = cos α

R.H.S =

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) =

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

Now,

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.

### Question 41. Prove that 1/(n+1) + 1/(n+2) +…..+ 1/2n > 13/24 for all natural numbers, n > 1.

**Solution:-**

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.

### Question 42. Given a_{1 }= 1/2(a_{0 }+ A/a_{0}), a_{2 }= 1/2(a_{1 }+ A/a_{1}) and a_{n+1 }= 1/2(a_{n }+ A/a_{n}), a, A > 0

### To prove:

**Solution:**

Let, P(n) =

Step 1:

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

LHS = (a

_{1}– √A) / (a_{1}+ √A)RHS =

= (a

_{1}– √A) / (a_{1}+ √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) =

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

_{k}^{2}+ A – 2a_{k}√A)/a_{k}) / (1/2(a_{k}^{2}+ A + 2a_{k}√A)/a_{k})= (a

_{k+1}– √A)^{2}/ (a_{k+1}+ √A)^{2}=

=

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.

### Question 43. Let P(n) be the statement: 2^{n} ≥ 3n. If P(r) is true then show that P(r + 1)is true, Do you conclude that P(n) is true for all n ∈ N?.

**Solution:**

Let P(n) = 2

^{n}≥ 3nStep 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 }≥ 3rStep 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 > 1or,2

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

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

^{r+1 }> 3(r + 1) for r >1So, 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.

### Question 44. Show by the principle of Mathematical Induction that the sum Sn of the n terms of the series 1^{2 }+ 2 × 2^{2 }+ 3^{2 }+ 2 × 4^{2 }+ 5^{2} + 2 × 6^{2 }+ 7^{2} +… is given by

**Solution:**

Let, P(n) = S

_{n}= 1^{2 }+ 2 × 2^{2 }+ 3^{2 }+ 2 × 4^{2 }+ 5^{2}+ 2 × 6^{2 }+ 7^{2}+… =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) = 1

^{2 }+ 2.2^{2 }+ 3^{2 }+ 2.4^{2 }+ 5^{2}=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 = 1

^{2 }+ 2.2^{2 }+ 3^{2 }+ 2.4^{2 }+ 5^{2}…. k^{2 }+ 2.(k + 1)^{2}= k

^{2}(k + 1)/2 + 2.(k + 1)^{2}= {(k

^{2}(k + 1) + 4(k + 1)^{2})}/2= (k + 1)(k + 2)

^{2}/2now, RHS = (k + 1)(k + 1 + 1)

^{2}/2= (k + 1)(k + 2)

^{2}/2So, 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 = 1

^{2 }+ 2.2^{2 }+ 3^{2 }+ 2.4^{2 }+ 5^{2}….+ 2. k^{2 }+ (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)/2RHS = (k + 1)

^{2}(k + 1 + 1)/2= (k + 1)

^{2}(k + 2)/2now, 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.

### Question 45. Prove that the number of subsets of a set containing n distinct elements is 2^{n} for all n ∈ N.

**Solution:**

Let, P(n): The number of subsets of a set containing n distinct elements = 2

^{n}, 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 = 2

^{1 }= 2now, 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 = {a

_{1}, a_{2}, a_{3}, a_{4},..…, 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: {}, {a

_{1}}, {a_{1}, a_{2}}, {a_{1}, a_{2}, a_{3}},…,{a_{1}, a_{2}, a_{3}, a_{4},…, a_{k}} andand the second collection: {b}, {a

_{1}, b}, {a_{1}, a_{2}, b}, {a_{1}, a_{2}, a_{3}, b},…,{a_{1}, a_{2}, a_{3}, a_{4},…,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. {a

_{1}, a_{2}, a_{3}, a_{4},…, a_{k}} = 2kAlso, 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.

### Question 46. A sequence a_{1}, a_{2}, a_{3},….. is defined by letting a_{1} = 3 and a_{k} = 7a_{k-1} for all numbers natural numbers k ≥ 2. Show that a_{n} = 3.7^{n-1} for all n ∈ N.

**Solution:**

Let P(n) be a

_{n }= 3.7^{n-1}for all n ∈ NStep 1:

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

so, a

_{1 }= 3.7^{1-1 }= 3So, 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.

### Question 47. A sequence x_{1}, x_{2}, x_{3},….. is defined by letting x_{1} = 2 and x_{k} = x_{k-1} /k for all numbers natural numbers k, k ≥ 2. Show that x_{n} = 2/n! for all n ∈ N.

**Solution:**

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) = x

_{1}= 2/1! = 2So, 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.

### Question 48. A sequence x_{0}, x_{1}, x_{2}, x_{3}, ……. is defined by letting x_{0 }= 5 and x_{k} = 4 + x_{k -1} for all natural number k. Show that x_{n} = 5 + 4_{n} for all n ∈ N using mathematical induction.

**Solution:-**

Let, P(n) = x

_{n}= 5 + 4n for all n ∈ NStep 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 + 4kStep 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.

### Question 49. Using principle of mathematical induction prove that

### √n < 1/√1 + 1/√2+ 1/√3 +…..+1/√n for all natural numbers n ≥2.

**Solution:**

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.

**Question 50. The distributive law from algebra states that for all real numbers, a**_{1} and a_{2}, we have c (a_{1} + a_{2}) = ca_{1} + ca_{2}. Use this law and mathematical induction to prove that, for all natural numbers, n ≥ 2, if c, a_{1}, a_{2},…,a_{n }are any real numbers, then c (a_{1}+a_{2}+…+a_{n}) = ca_{1}+ca_{2}+…+ca_{n}.

_{1}and a

_{2}, we have c (a

_{1}+ a

_{2}) = ca

_{1}+ ca

_{2}. Use this law and mathematical induction to prove that, for all natural numbers, n ≥ 2, if c, a

_{1}, a

_{2},…,a

_{n }are any real numbers, then c (a

_{1}+a

_{2}+…+a

_{n}) = ca

_{1}+ca

_{2}+…+ca

_{n}.

**Solution:**

Let, P(n) = c (a

_{1}+a_{2}+…+a_{n}) = ca_{1}+ca_{2}+…+ca_{n}, for all natural numbers, n ≥ 2.Step 1:

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

LHS = c(a

_{1}+ a_{2})RHS = c a

_{1}+ ca_{2}So, P(2) is true.

Step 2:

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

P(k) = c(a

_{1}+a_{2}+…+a_{k}) = ca_{1}+ca_{2}+…+ca_{k}Step 3:

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

LHS = c(a

_{1}+a_{2}+…+a_{k }+ a_{k+1})= c[(a

_{1}+a_{2}+…+a_{k}) + a_{k+1}]= c(a

_{1}+a_{2}+…+a_{k}) + ca_{k+1}= ca

_{1}+ca_{2}+…+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.