Mathematical induction

The Principles of Mathematical Induction

First principle of mathematical induction (weak mathematical induction)

The set of statements {P(n) : n ∈ N} is true for each natural number n ≥ m provided that:

1. P(m) is true.
2. P(k) is true for n = k, (where k ≥ m) ⟹ P(n) is true for n = k + 1.

Working Rule

Let there be a proposition or a mathematical statement namely P(n), involving a natural number n. In order to prove that P(n) is true for all natural numbers n ≥ m, we proceed as follows:

1. Verify that P(m) is true.
2. Assume that P(k) is true (where k ≥ m).
3. Prove that P(k + 1) is true.

Once step-3 is completed after 1 and 2, we are through. That is, P(n) is true for all natural numbers n ≥ m.

Second Principle of Mathematical Induction (strong mathematical induction)

The set of statements {P(n) : n ∈ N} is true for each natural number n ≥ m provided that:

1. P(m) and P(m + 1) are true.
2. P(n) is true for n ≤ k (where k ≥ m) ⟹ P(n) is true for n = k + 1.

This is also called extended principle of Mathematical Induction.

Working Rule

1. Verify that P(n) is true for n = m, n = m + 1.
2. Assume that P(n) is true for n ≤ k (where k ≥ m).
3. Prove that P(n) is true for n = k + 1.

Once step-3 is completed after 1 and 2, we are through. That is, P(n) is true for all natural numbers n ≥ m. (This method is to be used when P(n) can be expressed as a combination of P(n − 1) and P(n − 2). In case P(n) turns out to be a combination of P(n −1), P(n −2), and P(n −3), we verify for n = m + 2 also in step 1).

Illustration 1: For every natural number n, 11ⁿ⁺² + 12²ⁿ⁺¹ is divisible by

(A) 133

(B) 123

(C) 91

(D) none of these

Solution: (A). Let P(n) = 11ⁿ⁺² + 12²ⁿ⁺¹

P(1) = 11³ + 12³ = 3059 which is divisible by 133.

Let P(k) = 11^k+2 + 12^2k+1 be divisible by 133.

P(k+1) = 11^k+3 + 12^2k+3 = 11^k+2.11 + 12^2k+1.144

= 11.11^k+2 + (133 + 11) 12^2k+1

= 11[11^k+2 + 12^2k+1] + 133.12^2k+1

= 11.P(k) + 133.12^2k+1

P(K) is divisible by 133 also 133.12^2k+1 is divisible by 133. Hence, P (k + 1) is also divisible by 133. Hence, by mathematical induction, the result is true for all n.

Illustration 2: For every natural number n, 5^{2n + 2} –24n – 25 is divisible by

(A) 502

(B) 223

(C) 471

(D) 576

Solution: (D). Let P (n) = 5^{2n + 2} –24n –25

Then P (1) = 5⁴ –24 –25 = 576

Then P (1) is divisible by 576

Now assume that P (k) is divisible by 576, that is assume

P (k) = 5^{2k + 2} – 24k – 25 = 576m ……(1)  where m is a positive integer.

P (k + 1) = 5^{2(k + 1) + 2} – 24(k + 1) – 25 = 5².5^{2k + 2} – 24k – 49

= 25[5^{2k + 2} – 24k – 25] + 25.24k + 25.25 – 24k – 49 = 25P (k) + 576k + 576

= 25.576m + 576(k + 1) by (1) = 576(25m + k + 1)

Hence P (k + 1) is divisible by 576. Hence by mathematical induction, the result is true for all n.

Exercise 1:

(i) 7^2n-1 is divisible by, for all n ∈ N

(A) 8

(B) 7

(C) 11

(D) 13

(ii) 5.2ⁿ⁺¹ + 2ⁿ⁺⁴ + 2ⁿ⁺¹ is divisible by, for all n ∈ N

(A) 21

(B) 23

(C) 19

(D) none of these

(iii) 4ⁿ + 15n –1 is divisible by, for all n ∈ N

(A) 5

(B) 4

(C) 9

(D) 13

Illustration 3: The value of 1/(1.2.3) + 1/(2.3.4) + ... + 1/(n(n+1)(n+2)) is

(A) n(n+1)/2

(B) n(n+1)(2n+1)/6

(C) n(n+3)/(4(n+1)(n+2))

(D) none of these

Solution: (C).

Let P(n): 1/(1.2.3) + 1/(2.3.4) + ... + 1/(n(n+1)(n+2)) = n(n+3)/(4(n+1)(n+2))

P(1) = 1/(1.2.3) = 1/6 on the L.H.S; on the R.H.S = 1(1+3)/(4(1+1)(1+2)) = 1/6

∴ P(1) is true Let P(k) be true.

1/(1.2.3) + 1/(2.3.4) + ... + 1/(k(k+1)(k+2)) = k(k+3)/(4(k+1)(k+2))

P(k + 1) on the L.H.S.

= [1/(1.2.3) + 1/(2.3.4) + ... + 1/(k(k+1)(k+2))] + 1/((k+1)(k+2)(k+3))

The sum within the square brackets is P(k) which equals k(k+3)/(4(k+1)(k+2))

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

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

which is the desired R.H.S. of P(k+1). Hence, P(k+1) is true. Hence, by mathematical induction, the result is true for all n.

Illustration 4: The value of [1 - 1/2²][1 - 1/3²].....[1 - 1/(n+1)²] is, ∀ n ∈ N

(A) (n-1)/4

(B) (n+2)/(2n+2)

(C) n(n-1)/6

(D) (n+2)/(4(n+1))

Solution: (B).

Let P(n): [1 - 1/2²][1 - 1/3²].....[1 - 1/(n+1)²] = (n+2)/(2n+2)

L.H.S. of P(1) = [1 - 1/2²] = 3/4; R.H.S. = 3/4

Hence P(1) is true.

Assume that P(k) is true.

[1 - 1/2²][1 - 1/3²].....[1 - 1/(k+1)²] = (k+2)/(2k+2)

For P(k + 1), the L.H.S. becomes

[1 - 1/2²][1 - 1/3²].....[1 - 1/(k+1)²][1 - 1/(k+2)²]

=P(k)[1 - 1/(k+2)²] = (k+2)/(2k+2) × [(k+2)² - 1]/(k+2)²

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

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

∴ P(k+1) is true

Hence, by mathematical induction, the result is true for all n ∈ N

Use of Transitive Property

Suppose it is given F(n) > G(n) or F(n)/G(n) > 1

We have to prove that

F(n +1 ) > G(n + 1) or F(n+1)/G(n+1) > 1

or F(n+1)/G(n+1) ≥ F(n)/G(n) > 1

or F(n+1)/F(n) × G(n)/G(n+1) > 1.

Illustration 5: For all n ∈ N, ²ⁿCₙ

(A) < 4ⁿ

(B) > 4ⁿ

(C) > 2n + 1

(D) none of these

Solution: (A). To show ²ⁿCₙ < 4ⁿ

F(n) = 4ⁿ, G(n) = ²ⁿCₙ

F(n+1) = 4ⁿ⁺¹, G(n+1) = ²⁽ⁿ⁺¹⁾Cₙ₊₁

F(n+1)/F(n) × G(n)/G(n+1) = 4^(n+1)/4^n × ²ⁿCₙ/²⁽ⁿ⁺²⁾Cₙ₊₁ = 4 × (2n)!/(n!n!) × (n+1)!(n+1)!/((2n+2)!) = 2 × (n+1)/(2n+1) > 1   ( since 2n + 2 > 2n + 1 )

∴ F(n+1)/G(n+1) × G(n)/F(n) > 1 or F(n+1)/G(n+1) > F(n)/G(n) > 1 ∴ F(n+1) > G(n+1)

Illustration 6: For all natural numbers n greater than 1, the value of 1 + 1/4 + ..... + 1/n²

(A) ≥ 2 - 1/n

(B) ≤ 2 - 1/n

(C) < 2 - 1/n

(D) > 1/n

Solution: (C). For n = 2,

L.H.S = 1+ 1/4 = 5/4 and R.H.S. = 2- 1/2 = 3/2

5/4 < 3/2. Hence it holds for n = 2.

Assume the result to hold for n = k 1 + 1/4 + 1/9 + .... + 1/k² < 2 - 1/k

For n = k + 1, [1+ 1/4 + 1/9 + .... + 1/k²] + 1/(k+1)² < 2 - 1/k + 1/(k+1)²

Now, if we show that

2 - 1/k + 1/(k+1)² < 2 - 1/(k+1)

then we are through. 1/(k+1)² - 1/k + 1/(k+1) < 0

⟹ 1/k - 1/(k+1)² > 1/(k+1) ⟹ (k²+2k+1-k-k(k+1))/(k(k+1)²) > 0

= (k²-2k+1)/(k(k+1)²) = 1/(k(k+1)²) > 0.

Hence the result is true for n = k + 1. Hence, by mathematical induction, the result is true for all n.

Illustration 7: For what natural number n, the inequality 2ⁿ > 2n + 1 is valid ?

(A) n ≥ 3

(B) n > 1

(C) n = 2

(D) n > 5

Solution: (A). For n = 1, 2 < 2 + 1

n = 2, 4 < 5 n = 3, 8 > 7

so the result is true for n = 3

assume the result is true for n = k > 3, i.e.

2ᵏ > 2k + 1

So, P (k + 1) = 2^(k + 1) = 2.2ᵏ > 2.(2k + 1) = 4k + 2

= [2(k + 1) + 1] + (2k –1) > 2(k + 1) + 1 as 2k –1 > 0, since k > 3

∴ The result is true for n = k + 1

∴ By the principal of mathematical induction the result holds ∀ n ≥ 3.

Illustration 8: If x + y = a + b, x² + y² = a² + b², then xⁿ + yⁿ = aⁿ + bⁿ

(A) ∀ n ∈ N

(B) n ≥ 4

(C) n ≥ 3

(D) none of these

Solution: (A). Let P(n) ⟺ xⁿ + yⁿ = aⁿ + bⁿ

P(1) ⟺ x + y = a + b … (1)

P(2) ⟺ x² + y² = a² + b² … (2)

Hence P(1) and P(2) are true.

Assume the result to be true for n ≤ k.

∴ x^(k – 1) + y^(k – 1) = a^(k – 1) + b^(k – 1) and xᵏ + yᵏ = aᵏ + bᵏ

In order to prove that P(k + 1) is true, we write x^(k + 1) + y^(k + 1) = x(aᵏ + bᵏ – yᵏ) + y(aᵏ + bᵏ – xᵏ)

= (aᵏ + bᵏ) (x + y) – xy (x^(k – 1) + y^(k – 1))

= (aᵏ + bᵏ) (a + b) – xy (a^(k – 1) + b^(k – 1))

Now from (1) and (2) xy = ab

∴ x^(k + 1) + y^(k + 1) = a^(k + 1) + b^(k +1) which is the desired R.H.S. for P(k + 1).

Hence, by mathematical induction, the result is true for all n.

Illustration 9: If u₁ = cos θ, u₂ = cos 2θ and uₙ = 2uₙ₋₁ cos θ – uₙ₋₂ for n > 2. Then uₙ

(A) tan nθ

(B) cot nθ

(C) sin nθ

(D) cos nθ

Solution: (D). We have to prove that uₙ = cos nθ ∀ n ∈ N

Clearly u₁ = cos θ, u₂ = cos 2θ

So the result is true for n = 1 and n = 2 …(1)

Now assume the result for 2 < n ≤ k …(2)

Then uₖ₊₁ = 2uₖcos θ – uₖ₋₁ = 2 cos kθ cos θ – cos (k –1)θ

Using (2) = cos (k + 1)θ + cos (k –1)θ –cos (k –1)θ = cos (k + 1)θ …(3)

Hence the result is true for n = k + 1

Hence by mathematical induction the result is true for all n ∈ N.

Illustration 10: Let Iₘ = ∫₀^π (1 - cos mx)/(1 - cos x) dx, use mathematical induction, Iₘ

(A) (m-1)π/4, m = 0,1, 2, ...

(B) (m-1)π/3, m = 0,1, 2, ...

(C) mπ/2, m = 0, 1, 2…

(D) mπ, m = 0, 1, 2 ...

Solution: (D). I₁ = ∫₀^π (1 - cos x)/(1 - cos x) dx = ∫₀^π

1 dx = π I₂ = ∫₀^π (1 - cos 2x)/(1 - cos x) dx = ∫₀^π 2 dx = 2π

The result is true for m = 1, 2.  Let the result be true for m ≤ k.

Now, 2Iₖ – Iₖ₋₁ – Iₖ₊₁

= ∫₀^π [2(1 - cos kx) - (1 - cos(k-1)x) - (1 - cos(k+1)x)]/(1 - cos x) dx

= ∫₀^π [cos(k-1)x - cos kx - cos(k+1)x + cos kx]/(2sin²(x/2)) dx

= ∫₀^π [-2sin(kx/2)sin(x/2)]/(2sin²(x/2)) dx

= -∫₀^π cos kx dx = 0 ⟹ Iₖ₊₁ = 2Iₖ – Iₖ₋₁ = (k + 1)π

∴ The result is true for m = k + 1.

Hence, by mathematical induction, the result is true for all n.

Exercise 2:

For all n ∈ N, the value of

(i) 1.3 + 2.4 + 3.5 +...+ n.(n + 2) is equal to

(A) (1/6)n(n + 1)(2n + 7)

(B) (1/3)n(n + 1)(2n + 7)

(C) (1/6)n(n + 1)(2n + 7)

(D) (1/3)2n(n + 1)(n + 7)

(ii) 1/(2.5) + 1/(5.8) + 1/(8.11) + ... + 1/((3n-1)(3n+2)) is equal to

(A) n/(6n + 4)

(B) n/(6n + 4)

(C) n/(2n + 1)

(D) n/(2n + 1)

Solved Examples

Ques 1. For all n ∈ N, 1² + 2² + 3² + ……..+ n² is equal to

(A) (1/6)n(n – 1)(2n + 1)

(B) (1/6)n(n + 1)(2n + 1)

(C) (1/6)n(n + 1)(2n – 1)

(D) none of these

Sol. (B). Let P (n) be the statement given by

P (n) : 1² + 2² + 3² + …….. + n² = (1/6)n(n + 1)(2n + 1)

Step I: P (1) = 1² = (1/6) × 1 × 2 × 3 = 1

Hence P (1) is true.

Step II: P (m) be true, then

1² + 2² + 3² + …… + m² = (1/6)m(m + 1)(2m + 1)

we wish to show that P (m + 1) is true

1² + 2² + 3² + …… + m² + (m + 1)² = (1/6)m(m + 1)(2m + 1) + (m + 1)²

= (m + 1)/6 × [m(2m + 1) + 6(m + 1)]

= (m + 1)/6 × (2m² + m + 6m + 6) = (m + 1)(m + 1 + 1)(2(m + 1) + 1)/6.

So P (m + 1) is true.

Ques 2. (√3 + 1)^(2n) + (√3 – 1)^(2n) is

(A) negative rational number

(B) irrational number

(C) positive integer

(D) none of these

Sol. (C). (√3 + 1)^(2n) + (√3 – 1)^(2n)

= [(√3)^(2n) + ²ⁿC₁(√3)^(2n –1) + ²ⁿC₂(√3)^(2n – 2) + ….] + [(√3)^(2n) –²ⁿC₁(√3)^(2n –1) + …….]

= 2[(√3)^(2n) + ²ⁿC₂(√3)^(2n – 2) + ….]

= 2[3ⁿ + ²ⁿC₂3^(n –1) + ……] = positive integer.

Ques 3. For all n ∈ N, 3^(2n + 2) – 8n – 9 is divisible by

(A) 44

(B) 54

(C) 64

(D) none of these

Sol. (C). Given that 3^(2n + 2) – 8n – 9 = 3²(3^(2n)) – 8n – 9 = 9(9ⁿ) – 8n – 9

= 9(1 + 8)ⁿ – 8n – 9 = 9[1 + ⁿC₁8 + ⁿC₂8² + …..+ 8ⁿ] – 8n – 9

= 9 + 72n + 9(ⁿC₂8² + …… + 8ⁿ) – 8n – 9

= 64n + 64 × 9(ⁿC₂ + …….+ 8^(n –2))

= 64k, where k is an integer Hence 3^(2n + 2)– 8n – 9 is divisible by 64.

Ques 4. If 1/a + 1/b + 1/c = 1/(a+b+c), the value of (a + b + c)^(2n + 1), for n ∈ N is equal to

(A) a^(2n) + b^(2n + 1) + c^(2n)

(B) a^(2n + 1) + b^(2n + 1) + c^(2n + 1)

(C) a^(2n) + b^(2n) + c^(2n)

(D) none of these

Sol. (B). For n = 1,

= (a + b + c)³ = a³ + b³ + c³ + 3ab² + 3ac² + 3a²b + 3a²b + 3a²c + 3b²c + 3c²b + 6abc

= a³ + b³ + c³ + 3(a + b + c)(ab + bc + ca) – 3abc

= a³ + b³ + c³ + 3abc(a + b + c)(1/a + 1/b + 1/c) – 3abc = a³ + b³ + c³

Hence the result if true for n = 1

For n = 2,

= (a + b + c)⁵ = (a + b + c)³(a + b + c)² = (a³ + b³ + c³) (a² + b² + c²) + 2abc(a + b + c)²

= a⁵ + b⁵ + c⁵ + a²(b³ + c³) + b²(c³ + a³) + c²(a³ + b³) + 2abc(a + b + c)²

= a⁵ + b⁵ + c⁵ + [(ab + bc + ac)² – b²c²]a + [(ab + bc + ac)² – c²a²]b + [(ab + bc + ac)² – a²b²]c

= a⁵ + b⁵ + c⁵ + a²b²c²(1/a + 1/b + 1/c)²(a + b + c) – a²b²c²(1/a + 1/b + 1/c)

= a⁵ + b⁵ + c⁵

Hence the result if true for n = 2

Let the result is true for n ≤ m

For n = m + 1, we have (a + b + c)^(2m + 3) = (a + b + c)^(2m + 1)(a + b + c)²

= (a^(2m + 1) + b^(2m + 1) + c^(2m + 1))(a² + b² + c²) + 2abc(1/a + 1/b + 1/c).(a + b + c)^(2m + 1)

= a^(2m + 3) + b^(2m + 3) + c^(2m + 3) + a²(b^(2m + 1) + c^(2m + 1)) + b²(a^(2m + 1) + c^(2m + 1)) + c²(a^(2m + 1) + b^(2m + 1)) + 2abc (a + b + c)(a^(2m –1) + b^(2m –1) + c^(2m –1))

= a^(2m + 3) + b^(2m + 3) + c^(2m + 3) + (a²b² + b²c²)b^(2m –1) + (b²c² + a²c²)c^(2m –1) + (a²c² + a²b²)a^(2m –1) + 2abc(a + b + c)(a^(2m –1) + b^(2m –1) + c^(2m –1))

= a^(2m + 3) + b^(2m + 3) + c^(2m + 3) + [(ab + bc + ca)² – b²c²]a^(2m –1) + [(ab + bc + ca)² – c²a²]b^(2m –1) + [(ab + bc + ca)² – a²b²]c^(2m –1)

= a^(2m + 3) + b^(2m + 3) + c^(2m + 3) + a²b²c²(1/a + 1/b + 1/c)(a^(2m –1) + b^(2m –1) + c^(2m –1)) – a²b²c²(a^(2m –3) + b^(2m –3) + c^(2m – 3))

= a^(2m + 3) + b^(2m + 3) + c^(2m + 3) + a²b²c²(a + b + c)^(2m –3) – a²b²c²(a^(2m –3) + b^(2m –3) + c^(2m – 3))

= a^(2m + 3) + b^(2m + 3) + c^(2m + 3)

Hence the result is true for n = m + 1.

Hence by M.I the result is true for all n ∈ N.

Ques 5. For all n ∈ N, 1³ + 2³ + 3³ +....... + n³ is equal to

(A) [n(n + 1)/2]²

(B) [n(n + 1)/2]

(C) [n(n + 1)/2]²

(D) none of these

Sol. (C). Let P(n): 1³ + 2³ +....... + n³ = [n(n + 1)/2]²

Step I. When n = 1, L.H.S. = 1³ = 1 [when n = 1, L.H.S. has only one term] and R.H.S. = [1(1 + 1)/2]² = 1

∴ L.H.S. = R.H.S.

Hence P(1) is true …(A)

Step II. Let P(m) be true 1³ + 2³ + 3³ +........ + m³ = [m(m + 1)/2]² …(1)

To prove P(m + 1) is true

i.e. 1³ + 2³ +........ + m³ + (m + 1)³ = [(m + 1)(m + 2)/2]² …(2)

Adding (m + 1)³to both sides of (1), we get

1³ + 2³ + 3³ +....... + m³ + (m + 1)³ = [m(m + 1)]²/4 + (m + 1)³

= (m + 1)²[m² + 4(m + 1)]/4 = (m + 1)²(m² + 4m + 4)/4

= (m + 1)²(m + 2)²/4 = [(m + 1)(m + 2)/2]² …(3)

Hence P(m + 1) is true whenever P(m) is true …(B)

From (A) and (B) by the principle of induction it follows that P(n) is true for all natural numbers n.

Ques 6. For all n ∈ N, the value of 1²/(1.3) + 2²/(3.5) + ... + n²/((2n-1)(2n+1)), for n ≥ 1

(A) n(n+1)/(2(2n+1))

(B) n(n+1)/(2(2n+1))

(C) n(n+1)/(2(2n+1))

(D) none of these

Sol. (A). Let P(n): 1²/(1.3) + 2²/(3.5) + ... + n²/((2n-1)(2n+1)) = n(n+1)/(2(2n+1))

For n = 1, L.H.S is 1²/(1.3) = 1/3; R.H.S. = 1(2)/(2(3)) = 1/3 ∴ The result is true for n = 1.

Let us assume it to be true for n = k .i.e,

1²/(1.3) + 2²/(3.5) + ... + k²/((2k-1)(2k+1)) = k(k+1)/(2(2k+1))

Let us examine P(k+1). Then

1²/(1.3) + 2²/(3.5) + ... + k²/((2k-1)(2k+1)) + (k+1)²/((2k+1)(2k+3)) = P(k) + (k+1)²/((2k+1)(2k+3))

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

= (k+1)/(2k+1) × [k(2k+3) + 2(k+1)]/(2(2k+3)) = (k+1)/(2k+1) × [2k² + 5k + 2]/(2(2k+3))

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

P(k + 1) is true.

Hence, by mathematical induction, the result is true for all n.

Assignment

Ques. Let P(n) be a statement and let P(n) ⟹ P(n + 1) where n is natural number, then P(n) is true

(A) for all n

(B) for all n ≥ 1

(C) for all n ≥ m, m being a fixed positive integer

(D) nothing can be said.

Ques.  If x ≠ –1, then the statement (1 + x)ⁿ ≥ 1 + nx is true for

(A) all n ∈ N

(B) all n ≥ 1

(C) all n ≥ 1 provided x ≥ 0

(D) none of these

Ques.  If P(n) = 2 + 4 + 6 +……+ 2n, n∈N, then P(k) = k(k + 1) + 2 ⟹ P(k + 1) = (k + 1) (k + 2) + 2 for all k∈N. So we can conclude that P(n) = n(n + 1) + 2 for

(A) all n∈N

(B) n ≥ 1

(C) n ≥ 2

(D) nothing can be said.

Ques. x(xⁿ⁻¹ – naⁿ⁻¹) + aⁿ(n – 1) is divisible by (x – a)² for

(A) n ≥ 1

(B) n ≥ 2

(C) all n∈N

(D) none of these

Ques. The statement P(n) "1 × 1! + 2 × 2! + 3 × 3! +.......+ n × n! = (n + 1)! – 1" is

(A) true for all n ≥ 1

(B) not true for any n

(C) true for all n∈N

(D) none of these

Ques. A student was asked to prove a statement P(n) by method of induction. He proved that P(3) is true and that P(n) ⟹ P(n + 1) for all natural numbers n. On the basis of this he could conclude that P(n) is true

(A) for all n∈N

(B) for all n ≥ 3

(C) for no n

(D) none of these

Ques. The statement 2ⁿ + 2 ≤ 3ⁿ is true for

(A) all n∈N

(B) all n ≥ 2

(C) all n ≥ 3

(D) none of these

Ques. For all n ∈ N, the value of 1/(1.2.3.4) + 1/(2.3.4.5) + …. up to n terms is equal to

(A) 1/18 – 1/(3(n+1)(n+2)(n+3))

(B) 1/18 + 1/(3(n+1)(n+2)(n+3))

(C) 1/9 – 1/(3(n+1)(n+2)(n+3))

(D) none of these

Ques. If n is any odd positive integer, then n(n² – 1) is divisible by

(A) 11

(B) 5

(C) 24

(D) none of these

Ques. If n is even, then n (n²+20) is divisible by, ∀ n ∈ N

(A) 21

(B) 18

(C) 23

(D) 48

Ques. For all n ∈ N the value of 8 + 88 + 888 + …… + 88 ……. 8 (n digits) is equal to

(A) (8/9)(10ⁿ⁺¹ + 9n – 10)

(B) (8/81)(10ⁿ⁺¹ – 9n + 10)

(C) (8/9)(10ⁿ⁺¹ – 9n – 10)

(D) (8/81)(10ⁿ⁺¹ – 9n – 10)

Ques. For all n ∈ N, the value of cos a cos 2a cos 4a...cos (2ⁿ⁻¹a) is equal to

(A) cos²ⁿa/(2ⁿ sin a)

(B) sin²ⁿa/(2ⁿ cos a)

(C) cos²ⁿa/(2ⁿ cos a)

(D) sin²ⁿa/(2ⁿ sin a)

Ques. For all n ∈ N, (cos θ + i sin θ)ⁿ is equal to

(A) cos nθ + i sin nθ

(B) cos nθ – i sin nθ

(C) sin nθ + i sin nθ

(D) sin nθ + i cos nθ

Ques. The value of √(2 + √(2 + √(2 +...... √2......n times))) is equal to, ∀ n ∈ N

(A) 2cos(π/2ⁿ⁺¹)

(B) 2sin(π/2ⁿ⁺¹)

(C) 2cos(π/2ⁿ⁻¹)

(D) 2sin(π/2ⁿ⁻¹)

Ques. For all n ∈ N, 24ⁿ – 15n – 1 is divisible by

(A) 90

(B) 15

(C) 212

(D) 225

Answers to Assignment

1. D

2. C

3. D

4. C

5. C

6. B

7. C

8. A

9. C

10. D

11. D

12. D

13. A

14. A

15. D

Exercise Answers

Exercise 1:

(i) A - 7²ⁿ⁻¹ is divisible by 8

(ii) B - 5·2ⁿ⁺¹ + 2ⁿ⁺⁴ + 2ⁿ⁺¹ is divisible by 23

(iii) C - 4ⁿ + 15n –1 is divisible by 9

Exercise 2:

(i) A - 1.3 + 2.4 + 3.5 +...+ n.(n + 2) = (1/6)n(n + 1)(2n + 7)

(ii) B - 1/(2.5) + 1/(5.8) + ... + 1/((3n-1)(3n+2)) = n/(6n + 4)