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

  • Last Updated : 20 May, 2021

Question 33. Prove that n11/11 + n5/5 + n3/3 – 62/165n is true for all n ∈ N.

Solution:

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.

Question 34. Prove that (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.

 Solution:

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

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

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

= 1/(2mtan(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)+…..+ (1/2m)tan(x/2m) + (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.

Question 35. Prove that (1 – 1/22)(1 – 1/32)(1 – 1/42)……(1 – 1/n2) = (n + 1)/2n is true for all n ∈ N.

Solution:

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.

Question 36. Prove that (2n)!/22n(n!)2 ≤ 1/√(3n + 1) is true for all n ∈ N.

Solution:

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)!/22m(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)!/22m+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)!/22m) × (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.



Question 37. Prove that 1 + 1/4 + 1/9 + 1/16 + …..+ 1/n2 < 2 – 1/n for all n > 2, n ∈ N.

Solution:

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.

Question 38. Prove that x2n-1 + y2n-1 is divisible by x + y.

Solution:

Let, P(n) be   x2n-1 + y2n-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) = x2m-1 + y2m-1= λ(x + y)         ……….. (i)

Step 3:

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

i.e. ,  P(m + 1) = x2m+1 + y2m+1

= x2m+1 + y2m+1 – x2m-1. y2  +  x2m-1 . y2

= x2m-1(x2 – y2) + y2(x2m-1+y2m-1)

= (x + y)(x2m-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.



Question 39. Prove that sinx + sin3x + …+ sin(2n – 1)x = sin2nx/sinx is true for all n ∈ N.

Solution:

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.

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

Solution:

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.

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 a1 = 1/2(a0 + A/a0), a2 = 1/2(a1 + A/a1) and an+1 = 1/2(an + A/an), a, A > 0

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

Solution:

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 = (ak+1 – √A) / (ak+1 + √A)

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

= (1/2(ak2+ A – 2ak√A)/ak ) / (1/2(ak2+ A + 2ak√A)/ak )

=  (ak+1 – √A)2 / (ak+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.

Question 43. Let P(n) be the statement: 2n ≥ 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) = 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 2r ≥ 3r

Step 3:

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

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

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

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

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

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

or, 2r+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.

Question 44. Show by the principle of Mathematical Induction that the sum Sn of the n terms of the series 12 + 2 × 22 + 32 + 2 × 42 + 52 + 2 × 62 + 72 +… is given by 

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}

Solution:

Let, P(n) = Sn = 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.

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

Solution:

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 = 2k

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,..…, ak, 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,…, ak} and

and the second collection: {b}, {a1, b}, {a1, a2, b}, {a1, a2, a3, b},…,{a1, a2, a3, a4,…,ak, 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,…, ak} = 2k

Also, it follows that the second collection must have

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

So the total number of subsets of A = 2k + 2k = 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.

Question 46. A sequence a1, a2, a3,….. is defined by letting a1 = 3 and ak = 7ak-1 for all numbers natural numbers k ≥ 2. Show that an = 3.7n-1 for all n ∈ N.

Solution:

Let P(n) be an = 3.7n-1 for all n ∈ N

Step 1: 



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

so, a1 = 3.71-1 = 3

So, P(1) is true.

Step 2:

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

P(k) = ak = 3.7k-1

Step 3:

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

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

= 7.3.7k-1

= 3.7k-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 x1, x2, x3,….. is defined by letting x1 = 2 and xk = xk-1 /k for all numbers natural numbers k, k ≥ 2. Show that xn = 2/n! for all n ∈ N.

Solution:

Let, P(n) be xn = 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) = xk = 2/k! 

Step 3:

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

so,  xk+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 x0, x1, x2, x3, ……. is defined by letting x0 = 5 and xk = 4 + xk -1 for all natural number k. Show that xn = 5 + 4n for all n ∈ N using mathematical induction.

Solution:- 

Let, P(n) = xn = 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) = xk = 5 + 4k  

Step 3:

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

so, P(k + 1) = xk+1 = 4 + xk+1 -1

= 4 + xk

= 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, a1 and a2, we have c (a1 + a2) = ca1 + ca2. Use this law and mathematical induction to prove that, for all natural numbers, n ≥ 2, if c, a1, a2,…,an are any real numbers, then c (a1+a2+…+an) = ca1+ca2+…+can.

Solution:

Let, P(n) = c (a1+a2+…+an) = ca1+ca2+…+can, 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+…+ak) = ca1+ca2+…+cak

Step 3:

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

LHS = c(a1+a2+…+ak + ak+1)

= c[(a1+a2+…+ak) + ak+1]

= c(a1+a2+…+ak) + cak+1

= ca1+ca2+…+cak + cak+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.

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