Important Questions for CBSE Class 11 Maths Chapter 4 - Principle of Mathematical Induction
CBSE Class 11 Maths Chapter-4 Important Questions - Free PDF Download
4 Marks Questions
1. For every integer n, prove that
Ans. P(n) : 7n-3n is divisible by 4
For n=1
P(1) : 71 -31 =4 which is divisible by Thus, P(1) is true
Let P(k) be true
2. Prove that
Ans.
3. Prove that
Ans.
4. Prove
Ans.
5. Prove 1.2+2.3+3.4+--+n
Ans.
6. Prove (2n+7)<(n+3)2
Ans.
7. Prove
Ans.
8. Prove 1.2+2.22+3.23+…+n.2n = (n-1)2 n + 1+2
Ans.
c
9. Prove that 2.7n + 3.5n – 5 is divisible by 24
Ans. P(n) : 2.7n + 3.5n – 5 is divisible by 24
for n = 1
P (1) : 2.71 + 3.51 – 5 = 24 is divisible by 24
Hence result is true for n = 1
Let P (K) be true
P (K) : 2.7K + 3.5K – 5
(i)
we want to prove that P (K+!) is True whenever P (K) is true
2.7K+1 + 3.5K+1 – 5 = 2.7K .71 + 3.5K . 51 -5
Hence by P M I p (n) is true for all n N.
10. Prove that 41n – 14n is a multiple of 27
Ans. P (n) : 41n – 14n is a multiple of 27
for n = 1
P (1) : 411 – 14 = 27, which is a multiple of 27
Let P (K) be True
P (K) : 41K – 14K
(i)
we want to prove that result is true for n = K + 1
41K+1– 14k+1 = 41K.41-14K. 14
is a multiple of 27
Hence by PMI p (n) is true for ace n N.
11. Using induction, prove that 10n + 3.4n+2 + 5 is divisible by 9 .
Ans. P (n) : 10n + 3.4n+2 + 5 is divisible by 9
For n = 1
p (1) : 101 + 3.41+2 + 5=207, divisible by 9
Hence result is true for n = 1
Let p (K) be true
p (K) : 10K +3.4K+2 + 5 is divisible by 9
we want to prove that result is true for n = K + 1
10(K+1) +3.4K+1+2+5=10K+1+3.4K+3+5
which is divisible by 9.
12. Prove that 12 + 22 + 32 + ---+ n2 =
Ans. Let P(n) : 12 + 22 + 32 + -- +n2 =
for n = 1
Thus P (K+1) is true, whenever P (K) is true.
Hence, from PMI, the statement P (n) is true for all natural no. n.
13. Prove that 1+3+32 + -- + 3n-1 =
Ans. Let
P (n) : 1 + 3 + 32 + -- + 3n-1 = for n = 1
P (1) 31-1 =
which is true
Let P (K) be true
P (K) : 1 + 3 + 32 + ---- + 3K-1 =
we want to prove that P (K+1) is true
Hence p (K+1) is true whenever p (K) is True
14. By induction, prove that 12 + 22 + 32 + -- + n2 >
Ans. Let P (n) : 12 + 22 + 32 + --- + n2 >
for n = 1
12 > which is true
Let P (K) be true
P (K) : 12 + 22 + 32 + -- + K2 >
we want to prove that P (K + 1) is true
P (K+1) : 12 + 22 + --- + (K+1)2
Hence by PMI P (n) is true
15. Prove by PMI (ab)n = an bn
Ans. Let P (n) : (ab)n = an bn
for n = 1
ab = ab which is true
Let P (K) be true
(ab)K = aK bK (1)
we want to prove that P (K+1) is true
(ab)K+1 = aK+1. bK+1
L.H.S = (ab)K+1
16. Prove by PMI a + ar + ar2 + ---- + arn-1 =
Ans. Let P (n) : a + ar + ar2 + -- + arn-1 =
for n = 1
P (1) = a = a which is true
Let P (K) be true
P (K) : a + ar + ar2 + -- + arK-1 =
we want to prove that
P (K+1) : a + ar + ar2 + -- + arK =
L.H.S
Thus P (K+1) is true whenever P(K) is true
Hence by PMI P(n) is true for all n N
17. Prove that x2n – y2n is divisible by x + y.
Ans. P (n) : x2n – y2n is divisible by x + y
for n = 1
p (1) : x2 – y2 = (x – y) (x + y), which is divisible by x + y
Hence result is true for n = 1
Let P (K) be true
p (K) : x2K – y2K is divisible by x + y
18. Prove that n (n + 1) (2n + 1) is divisible by 6.
Ans.P (n) : n (n+1) (2n+1) is divisible by 6 for n = 1
P (1) : (1) (2) (3) = 6 is divisible by 6
Hence result is true for n = 1
Let P (K) be true
P (K) : K (K+1) (2K+1) is divisible by 6
19. Show that 23n – 1 is divisible by 7 .
Ans.P (n) : 23n – 1 is divisible by 7
for n = 1
P (1) : 23 – 1 = 7 which is divisible by 7
Let P (K) be true
P (K) : 23K – 1 is divisible by 7
s
Thus P (K+1) is true
Hence by P.M.I P (n) is true
20. Prove by P M I.
1. 2. 3 + 2. 3. 4 + --- + n (n + 1) (n + 2)
=
Ans.Let P (n) : 1. 2. 3 + 2. 3. 4 + --- + n (n+1) (n+2)
=
For n = 1
P (1) = 1 (2) (3) =
P (1) = 6 = 6 which is true
Let P (K) be true
P (K) : 1. 2. 3 + 2. 3. 4 + -- + K (K+1) (K+2)
=
we want to prove that
P (K+1) n: 1. 2. 3 + 2. 3. 4 + -- + (K+1) (K+2) (K+3) =
L.H.S = 1. 2. 3 + 2. 3. 4 + -- + K (K+1) (K+2) + (K+1) (K+2) (K+3)
Thus P (K+1) is true whenever P(K) is true.
21. Prove that
Ans.P(n) :
For n = 1
P (1) =
Let P (K) be true
Thus P (K+1) is for whenever P (K) is true.
22. Show that the sum of the first n odd natural no is n2.
Ans.Let P (n) : 1 + 3 + 5 + --- + (2n-1) = n2
For n = 1
P (1) = 1 = 1 which is true
Let P (K) be true
P (K) : 1 + 3 + 5 + --- + (2K-1) = K2 (1)
we want to prove that P (K+1) is true
P (K+1) : 1 + 3 + 5 + --- + (2K+1) = (K+1)2
L.H.S =
Thus P (K+1) is true whenever P(K) is true.
Hence by PMI, P(n) is true for all n N.
23. Prove by P M I
Ans.P (n) :
For n = 1
P (1) : 13 = 13 which is true
Let P (K) be true
Thus P (K+1) is true whenever P (K) is true.
24. Prove.
Ans.P (n) :
For n = 1
P (1) : 4 = 4 which is true
Let P (K) be true
P (K) :
We want to power that P (K+1) is true
Thus P (K+1) is true whenever P (K) is true.
25. Prove that 32n+2 – 8n – 9 is divisible by 8
Ans.P(n) : 32n+2 – 8n – 9 is divisible by 8
For n = 1
P (1) : 32+2 – 8 – 9 = 64
which is divisible by 8
Hence result is true for n = 1
Let P (K) be true
P (K) : 32K+2 – 8K – 9 is divisible by 8
( from i)
26. Prove by PMI.
xn-yn is divisible by (x-y) whenever x-y 0
Ans.P (n) : xn-yn is divisible by (x-y)
For n = 1
P (1) : x – y is divisible by (x – y)
Let P (K) be true
P (K) : xK – yK is divisible by (x – y)
( from i)
27. Prove (x2n-1) is divisible by (x-1).
Ans.P (n) : (x2n-1) is divisible by (x-1).
For n = 1
P (1) : (x2 – 1) = (x – 1) (x + 1)
which is divisible by (x – 1)
Let P (K) be true
28. Prove
Ans.P (n) :
we want to prove that p (k + 1) is true
29. Prove 1.3 + 3.5 + 5.7 + -- + (2n – 1) (2n + 1)
=
Ans.Let p (n) : 1.3 + 3.5 + -- + (2n-1) (2n+1)
=
For n = 1
P (1) = (1) (3) =
P(1) = 3 = 3 Hence p (1) is true
Let (k) be true
P(k) : 1.3 + 3.5 + -- + (2k - 1) (2K + 1) =
we want to prove that p (k+1) is true
p (k+1) : 1.3 + 3.5 + -- + (2k+1) (2k+3) =
L. H. S
Thus p (k+1) is true whenever p (k) is true.
30. Prove by PMI
3.22 + 32.23 + 33.24 + -- + 3n. 2n+1 = .
Ans.Let p (n) : 3.22 + 32.23 + 33.24 + --- + 3n. 2n+1 =
For n = 1
Thus p(k+1) is truewhenever p(k) is true.
31. Prove 1.3 + 2.32 + 3.33 + --- + n.3n =
Ans.P (n) : 1.3 + 2.32 + 3.33 + --- + n.3n =
For n = 1
P (1) : 1. 31 =
Thus p (k+1) is true whenever p(k) is true.
32. Prove
Ans. P(n) :
For n = 1
Thus p (k+1) is true whenever p (k) is true
Hence p (n) is true for all n N.
33. The sum of the cubes of three consecutive natural no. is divisible by 9.
Ans.P(n) is divisible by 9
For n = 1
P (1) : 1 + 8 + 9 = 18
which is divisible by 9
Let p (k) be true
34. Prove that 12n + 25n-1 is divisible by 13
Ans.P(n) : 12n + 25n-1 is divisible by 13
For n = 1
P(1) : 12 + (25)0 = 13
which is divisible by 13
Let p (k) be true
P(k) : 12k + 25k-1 is divisible by 13
which is divisible by 13.
35. Prove 11n+2 + 122n+1 is divisible by 133.
Ans. P(n) : 11n+2 + 122n+1 is divisible by 133.
For n = 1
P(1) : 113 + 123 = 3059
which is divisible by 133
Let p (k) be true
36. Prove 13 + 23 + 33 + --- + n3 =
Ans. P(n) : 13 + 23 + 33 + --- + n3 =
For n = 1
Thus p(k+1) is true whenever p(k) is true.
37. Prove (a)+ (a + d) + (a + 2d) + -- + [a + (n – 1)d] =
Ans. P(n) : (a)+ (a + d) + (a + 2d) + -- + [a + (n – 1)d] =
For n = 1
p(1) : a + (1-1) d = 2a + (1-1) d = a
which is true
Let p (k) be true
38. Prove that 2n > n positive integers n.
Ans. Let p (n) : 2n > n
For n = 1
P (1) : 21 >1
Which is true
Let p (k) be true
P (k) : 2k > k (1)
we want to prove that p (k+1) is true
2k> k by (1)
39. Prove
Ans. P(n) :
For n = 1
40. Prove
Ans.P(n) :
For n = 1