Students can download 11th Business Maths Chapter 2 Algebra Ex 2.5 Questions and Answers, Notes, Samcheer Kalvi 11th Business Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

## Tamilnadu Samacheer Kalvi 11th Business Maths Solutions Chapter 2 Algebra Ex 2.5

### Samacheer Kalvi 11th Business Maths Algebra Ex 2.5 Text Book Back Questions and Answers

By the principle of mathematical induction, prove the following:

Question 1.

1^{3} + 2^{3} + 3^{3} + ….. + n^{3} = \(\frac{n^{2}(n+1)^{2}}{4}\) for all x ∈ N.

Solution:

Let P(n) be the statement 1^{3} + 2^{3} + …… + n^{3} = \(\frac{n^{2}(n+1)^{2}}{4}\) for all n ∈ N.

i.e., p(n) = 1^{3} + 2^{3} + …… + n^{3} = \(\frac{n^{2}(n+1)^{2}}{4}\), for all n ∈ N

Put n = 1

LHS = 1^{3} = 1

RHS = \(\frac{1^{2}(1+1)^{2}}{4}\)

= \(\frac{1 \times 2^{2}}{4}\)

= \(\frac{4}{4}\)

= 1

∴ P(1) is true.

Assume that P(n) is true n = k

P(k): 1^{3} + 2^{3} + …… + k^{3} = \(\frac{k^{2}(k+1)^{2}}{4}\)

To prove P(k + 1) is true.

i.e., to prove 1^{3} + 2^{3} + ……. + k^{3} + (k + 1)^{3} = \(\frac{(k+1)^{2}((k+1)+1)^{2}}{4}=\frac{(k+1)^{2}(k+2)^{2}}{4}\)

Consider 1^{3} + 2^{3} + …… + k^{3} + (k + 1)^{3} = \(\frac{k^{2}(k+1)^{2}}{4}\) + (k + 1)^{3}

= (k + 1)^{2} [\(\frac{k^{2}}{4}\) + (k + 1)]

= (k + 1)^{2} \(\left[\frac{k^{2}+4(k+1)}{4}\right]\)

= \(\frac{(k+1)^{2}(k+2)^{2}}{4}\)

⇒ P(k + 1) is true, whenever P(k) is true.

Hence, by the principle of mathematical induction P(n) is true for all n ∈ N.

Question 2.

1.2 + 2.3 + 3.4 + …… + n(n + 1) = \(\frac{n(n+1)(n+2)}{3}\), for all n ∈ N.

Solution:

Let P(n) denote the statement

1.2 + 2.3 + 3.4 + …… + n(n + 1) = \(\frac{n(n+1)(n+2)}{3}\)

Put n = 1

LHS = 1(1 + 1) = 2

RHS = \(\frac{1(1+1)(1+2)}{3}=\frac{1(2)(3)}{3}\) = 2

∴ P(1) is true.

Now assume that the statement be true for n = k

(i.e.,) assume P(k) be true

(i.e.,) assume 1.2 + 2.3 + 3.4 + …… + k(k + 1) = \(\frac{k(k+1)(k+2)}{3}\) be true

To prove: P(k + 1) is true

(i.e.,) to prove: 1.2 + 2.3 + 3.4 + …… + k(k + 1) + (k + 1) (k + 2) = \(\frac{(k+1)(k+2)(k+3)}{3}\)

Consider 1.2 + 2.3 + 3.4 + ……. + k(k + 1) + (k + 1) (k + 2)

= [1.2 + 23 + …… + k(k + 1)] + (k + 1) (k + 2)

= \(\frac{k(k+1)(k+2)}{3}\) + (k + 1) (k + 2)

= \(\frac{k(k+1)(k+2)+3(k+1)(k+2)}{3}\)

= \(\frac{(k+1)(k+2)(k+3)}{3}\)

∴ P(k + 1) is true.

Thus if P(k) is true, P(k + 1) is true.

By the principle of Mathematical ‘induction, P(n) is true for all n ∈ N.

1.2 + 2.3 + 3.4 + …… + n(n + 1) = \(\frac{n(n+1)(n+2)}{3}\)

Question 3.

4 + 8 + 12 + ……. + 4n = 2n(n + 1), for all n ∈ N.

Solution:

Let P(n) denote the statement 4 + 8 + …….. + 4n = 2n(n + 1)

i.e., P(n) : 4 + 8 + 12 + … + 4n = 2n(n + 1)

Put n = 1,

P(1): LHS = 4

RHS = 2 (1)(1 + 1) = 4

P(1) is true.

Assume that P(n) is true for n = k

P(k): 4 + 8 + 12 + ……. + 4k = 2k(k + 1)

To prove P(k + 1)

i.e., to prove 4 + 8 + 12 + ……. + 4k + 4(k + 1) = 2(k + 1) (k + 1 + 1)

4 + 8 + 12 + …… + 4k + (4k + 4) = 2(k + 1) (k + 2)

Consider, 4 + 8 + 12 + …….. + 4k + (4k + 4) = 2k(k + 1) + (4k + 4)

= 2k(k + 1) + 4(k + 1)

= 2k^{2} + 2k + 4k + 4

= 2k^{2} + 6k + 4

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

P(k + 1) is also true.

∴ By Mathematical Induction, P(n) for all value n ∈ N.

Question 4.

1 + 4 + 7 + ……. + (3n – 2) = \(\frac{n(3 n-1)}{2}\) for all n ∈ N.

Solution:

Let P(n) : 1 + 4 + 7 + ……. + (3n – 2) = \(\frac{n(3 n-1)}{2}\)

Put n = 1,

LHS = 1

RHS = \(\frac{1(3-1)}{2}\) = 1

∴ P(1) is true.

Assume P(k) is true for n = k

P(k): 1 + 4 + 7 + ……. + (3k – 2) = \(\frac{k(3 k-1)}{2}\)

To prove P(k + 1) is true, i.e., to prove

1 + 4 + 7 + ……. + (3k – 2) + (3(k + 1) – 2) = \(\frac{(k+1)(3(k+1)-1)}{2}\)

1 + 4 + 7 + ……. + (3k – 2) + (3k + 3 – 2) = \(\frac{(k+1)(3 k+2)}{2}\)

1 + 4 + 7 + …… + (3k + 1) = \(\frac{(k+1)(3 k+2)}{2}\)

1 + 4 + 7 + …… + (3k – 2) + (3k + 1) = \(\frac{k(3 k-1)}{2}\) + (3k + 1)

= \(\frac{k(3 k-1)+2(3 k+1)}{2}\)

= \(\frac{3 k^{2}-k+6 k+2}{2}\)

= \(\frac{3 k^{2}+5 k+2}{2}\)

= \(\frac{(k+1)(3 k+2)}{2}\)

∴ P(k + 1) is true whenever P(k) is true.

∴ By the Principle of Mathematical Induction, P(n) is true for all n ∈ N.

Question 5

3^{2n} – 1 is divisible by 8, for all n ∈ N.

Solution:

Let P(n) denote the statement 3^{2n} – 1 is divisible by 8 for all n ∈ N

Put n = 1

P(1) is the statement 3^{2(1)} – 1 = 3^{2} – 1 = 9 – 1 = 8, which is divisible by 8

∴ P(1) is true.

Assume that P(k) is true for n = k.

i.e., 3^{2k} – 1 is divisible by 8 to be true.

Let 3^{2k} – 1 = 8m

To prove P(k + 1) is true.

i.e., to prove 3^{2(k+1)} – 1 is divisible by 8

Consider 3^{2(k+1)} – 1 = 3^{2k+2} – 1

= 3^{2k} . 3^{2} – 1

= 3^{2k} (9) – 1

= 3^{2k} (8 + 1) – 1

= 3^{2k} × 8 + 3^{2k} × 1 – 1

= 3^{2k} (8) + 3^{2k} – 1

= 3^{2k} (8) + 8m (∵ 3^{2k} – 1 = 8m)

= 8(3^{2k} + m), which is divisible by 8.

∴ P(k + 1) is true wherever P(k) is true.

∴ By principle of Mathematical Induction, P(n) is true for all n ∈ N.

Question 6.

a^{n} – b^{n} is divisible by a – b, for all n ∈ N.

Solution:

Let P(n) denote the statement a^{n} – b^{n} is divisible by a – b.

Put n = 1. Then P(1) is the statement: a^{1} – b^{1} = a – b is divisible by a – b

∴ P(1) is true. Now assume that the statement be true for n = k

(i.e.,) assume P(k) be true, (i.e.,) a^{k} – b^{k} is divisible by (a – b) be true.

⇒ \(\frac{a^{k}-b^{k}}{a-b}\) = m (say) where m ∈ N

⇒ a^{k} – b^{k} = m(a – b)

⇒ a^{k} = b^{k} + m(a – b) ……. (1)

Now to prove P(k + 1) is true, (i.e.,) to prove: a^{k+1} – b^{k+1} is divisible by a – b

Consider a^{k+1} – b^{k+1} = a^{k} . a – b^{k} . b

= [b^{k} + m(a – b)] a – b^{k} . b [∵ a^{k} = bm + k(a – b)]

= b^{k} . a + am(a – b) – b^{k} . b

= b^{k} . a – b^{k} . b + am(a – b)

= b^{k}(a – b) + am(a – b)

= (a – b) (b^{k} + am) is divisible by (a – b)

∴ P(k + 1) is true.

By the principle of Mathematical induction. P(n) is true for all n ∈ N.

∴ a^{n} – b^{n} is divisible by a – b for n ∈ N.

Question 7.

5^{2n} – 1 is divisible by 24, for all n ∈ N.

Solution:

Let P(n) be the proposition that 5^{2n} – 1 is divisible by 24.

For n = 1, P(1) is: 5^{2} – 1 = 25 – 1 = 24, 24 is divisible by 24.

Assume that P(k) is true.

i.e., 5^{2k} – 1 is divisible by 24

Let 5^{2k} – 1 = 24m

To prove P(k + 1) is true.

i.e., to prove 5^{2(k+1)} – 1 is divisible by 24.

P(k): 5^{2k} – 1 is divisible by 24.

P(k + 1) = 5^{2(k+1)} – 1

= 5^{2k} . 5^{2} – 1

= 5^{2k} (25) – 1

= 5^{2k} (24 + 1) – 1

= 24 . 5^{2k} + 5^{2k} – 1

= 24 . 5^{2k} + 24m

= 24 [5^{2k} + 24]

which is divisible by 24 ⇒ P(k + 1) is also true.

Hence by mathematical induction, P(n) is true for all values n ∈ N.

Question 8.

n(n + 1) (n + 2) is divisible by 6, for all n ∈ N.

Solution:

P(n): n(n + 1) (n + 2) is divisible by 6.

P(1): 1 (2) (3) = 6 is divisible by 6

∴ P(1) is true.

Let us assume that P(k) is true for n = k

That is, k (k + 1) (k + 2) = 6m for some m

To prove P(k + 1) is true i.e. to prove (k + 1) (k + 2)(k + 3) is divisible by 6.

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

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

= 6m + 3(k + 1)(k + 2)

In the second term either k + 1 or k + 2 will be even, whatever be the value of k.

Hence second term is also divisible by 6.

∴ P (k + 1) is also true whenever P(k) is true.

By Mathematical Induction P (n) is true for all values of n.

Question 9.

2^{n} > n, for all n ∈ N.

Solution:

Let P(n) denote the statement 2^{n} > n for all n ∈ N

i.e., P(n): 2^{n} > n for n ≥ 1

Put n = 1, P(1): 2^{1} > 1 which is true.

Assume that P(k) is true for n = k

i.e., 2^{k} > k for k ≥ 1

To prove P(k + 1) is true.

i.e., to prove 2^{k+1} > k + 1 for k ≥ 1

Since 2^{k} > k

Multiply both sides by 2

2 . 2^{k} > 2k

2^{k+1} > k + k

i.e., 2^{k+1} > k + 1 (∵ k ≥ 1)

∴ P(k + 1) is true whenever P(k) is true.

∴ By principal of mathematical induction P(n) is true for all n ∈ N.