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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.
13 + 23 + 33 + ….. + n3 = \(\frac{n^{2}(n+1)^{2}}{4}\) for all x ∈ N.
Solution:
Let P(n) be the statement 13 + 23 + …… + n3 = \(\frac{n^{2}(n+1)^{2}}{4}\) for all n ∈ N.
i.e., p(n) = 13 + 23 + …… + n3 = \(\frac{n^{2}(n+1)^{2}}{4}\), for all n ∈ N
Put n = 1
LHS = 13 = 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): 13 + 23 + …… + k3 = \(\frac{k^{2}(k+1)^{2}}{4}\)
To prove P(k + 1) is true.
i.e., to prove 13 + 23 + ……. + k3 + (k + 1)3 = \(\frac{(k+1)^{2}((k+1)+1)^{2}}{4}=\frac{(k+1)^{2}(k+2)^{2}}{4}\)
Consider 13 + 23 + …… + k3 + (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)
= 2k2 + 2k + 4k + 4
= 2k2 + 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
32n – 1 is divisible by 8, for all n ∈ N.
Solution:
Let P(n) denote the statement 32n – 1 is divisible by 8 for all n ∈ N
Put n = 1
P(1) is the statement 32(1) – 1 = 32 – 1 = 9 – 1 = 8, which is divisible by 8
∴ P(1) is true.
Assume that P(k) is true for n = k.
i.e., 32k – 1 is divisible by 8 to be true.
Let 32k – 1 = 8m
To prove P(k + 1) is true.
i.e., to prove 32(k+1) – 1 is divisible by 8
Consider 32(k+1) – 1 = 32k+2 – 1
= 32k . 32 – 1
= 32k (9) – 1
= 32k (8 + 1) – 1
= 32k × 8 + 32k × 1 – 1
= 32k (8) + 32k – 1
= 32k (8) + 8m (∵ 32k – 1 = 8m)
= 8(32k + 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.
an – bn is divisible by a – b, for all n ∈ N.
Solution:
Let P(n) denote the statement an – bn is divisible by a – b.
Put n = 1. Then P(1) is the statement: a1 – b1 = 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.,) ak – bk is divisible by (a – b) be true.
⇒ \(\frac{a^{k}-b^{k}}{a-b}\) = m (say) where m ∈ N
⇒ ak – bk = m(a – b)
⇒ ak = bk + m(a – b) ……. (1)
Now to prove P(k + 1) is true, (i.e.,) to prove: ak+1 – bk+1 is divisible by a – b
Consider ak+1 – bk+1 = ak . a – bk . b
= [bk + m(a – b)] a – bk . b [∵ ak = bm + k(a – b)]
= bk . a + am(a – b) – bk . b
= bk . a – bk . b + am(a – b)
= bk(a – b) + am(a – b)
= (a – b) (bk + 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.
∴ an – bn is divisible by a – b for n ∈ N.
Question 7.
52n – 1 is divisible by 24, for all n ∈ N.
Solution:
Let P(n) be the proposition that 52n – 1 is divisible by 24.
For n = 1, P(1) is: 52 – 1 = 25 – 1 = 24, 24 is divisible by 24.
Assume that P(k) is true.
i.e., 52k – 1 is divisible by 24
Let 52k – 1 = 24m
To prove P(k + 1) is true.
i.e., to prove 52(k+1) – 1 is divisible by 24.
P(k): 52k – 1 is divisible by 24.
P(k + 1) = 52(k+1) – 1
= 52k . 52 – 1
= 52k (25) – 1
= 52k (24 + 1) – 1
= 24 . 52k + 52k – 1
= 24 . 52k + 24m
= 24 [52k + 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.
2n > n, for all n ∈ N.
Solution:
Let P(n) denote the statement 2n > n for all n ∈ N
i.e., P(n): 2n > n for n ≥ 1
Put n = 1, P(1): 21 > 1 which is true.
Assume that P(k) is true for n = k
i.e., 2k > k for k ≥ 1
To prove P(k + 1) is true.
i.e., to prove 2k+1 > k + 1 for k ≥ 1
Since 2k > k
Multiply both sides by 2
2 . 2k > 2k
2k+1 > k + k
i.e., 2k+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.