Principle of Mathematical Induction Practice Questions - DPP for JEE

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? P.M.I cannot be applied
Let S(k) is true, i.e.
5. (b) Check through option, the condition  is satisfied for
n > 2
6. (a) Putting n = 2 in 3
2n
 –2n + 1 then,
 which is divisibleby 2.
7. (a) The product of two consecutive numbers is always even.
8. (b) a
1
 = < 7. Let a
m
 < 7
Then a
m
 
+ 1
 =   a
2
? m + 1
= 7 + a
m
 < 7 + 7 < 14.
 a
m + 1
 < < 7;  So by the principle of mathematical
induction a
n
 < 7 n. 
9. (c) Check through option, the condition is true when .
10. (d) Let P(n) : 
For n = 2,
P(2) :  ? 
which is true.
Let for n = m = 2, P(m) is true.
i.e.
Now,  = 
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? P.M.I cannot be applied
Let S(k) is true, i.e.
5. (b) Check through option, the condition  is satisfied for
n > 2
6. (a) Putting n = 2 in 3
2n
 –2n + 1 then,
 which is divisibleby 2.
7. (a) The product of two consecutive numbers is always even.
8. (b) a
1
 = < 7. Let a
m
 < 7
Then a
m
 
+ 1
 =   a
2
? m + 1
= 7 + a
m
 < 7 + 7 < 14.
 a
m + 1
 < < 7;  So by the principle of mathematical
induction a
n
 < 7 n. 
9. (c) Check through option, the condition is true when .
10. (d) Let P(n) : 
For n = 2,
P(2) :  ? 
which is true.
Let for n = m = 2, P(m) is true.
i.e.
Now,  = 
< 
= 
= 
< 
Hence, for n = 2, P(n) is true.
11. (a) x
2n–1
 + y
2n–1
 is always contain equal odd power. So it is always
divisible by x + y .
12. (a) It can be proved with the help of mathematical induction that  <
a(n) = n.
? < a(200) ? a(200) > 100 and
a(100) = 100.
13. (d) Let the given statement be P (n), then
P (1) ? 2
1
 > 1
2
 which is true
P (2) ? 2
2
 > 2
2
 which is false
P (3) ? 2
3
 > 3
2
 which is false
P (4) ? 2
4
 > 4
2
 which is false
P (5) ? 2
5
 > 5
2
 which is true
P (6) ? 2
6
 > 6
2
 which is true
?   P (n) is true when n = 5
14. (a) Let P (n) = n (n
2
 – 1) then
P(1) = 1 (0) = 0 which is divisible by every n ? N
P (3) = 3 (8) = 24 which is divisible by 24 and 8
P (5) = 5 (24) = 120 which is divisible by 24 and 8
Hence P (n) is divisible by 24.
15. (c) Let n = 1, then option (a), (b) and (d) eliminated. Only option (c)
satisfied.
16. (a) P(n) : 2.7
n
 + 3.5
n
 – 5 is divisible by 24.
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? P.M.I cannot be applied
Let S(k) is true, i.e.
5. (b) Check through option, the condition  is satisfied for
n > 2
6. (a) Putting n = 2 in 3
2n
 –2n + 1 then,
 which is divisibleby 2.
7. (a) The product of two consecutive numbers is always even.
8. (b) a
1
 = < 7. Let a
m
 < 7
Then a
m
 
+ 1
 =   a
2
? m + 1
= 7 + a
m
 < 7 + 7 < 14.
 a
m + 1
 < < 7;  So by the principle of mathematical
induction a
n
 < 7 n. 
9. (c) Check through option, the condition is true when .
10. (d) Let P(n) : 
For n = 2,
P(2) :  ? 
which is true.
Let for n = m = 2, P(m) is true.
i.e.
Now,  = 
< 
= 
= 
< 
Hence, for n = 2, P(n) is true.
11. (a) x
2n–1
 + y
2n–1
 is always contain equal odd power. So it is always
divisible by x + y .
12. (a) It can be proved with the help of mathematical induction that  <
a(n) = n.
? < a(200) ? a(200) > 100 and
a(100) = 100.
13. (d) Let the given statement be P (n), then
P (1) ? 2
1
 > 1
2
 which is true
P (2) ? 2
2
 > 2
2
 which is false
P (3) ? 2
3
 > 3
2
 which is false
P (4) ? 2
4
 > 4
2
 which is false
P (5) ? 2
5
 > 5
2
 which is true
P (6) ? 2
6
 > 6
2
 which is true
?   P (n) is true when n = 5
14. (a) Let P (n) = n (n
2
 – 1) then
P(1) = 1 (0) = 0 which is divisible by every n ? N
P (3) = 3 (8) = 24 which is divisible by 24 and 8
P (5) = 5 (24) = 120 which is divisible by 24 and 8
Hence P (n) is divisible by 24.
15. (c) Let n = 1, then option (a), (b) and (d) eliminated. Only option (c)
satisfied.
16. (a) P(n) : 2.7
n
 + 3.5
n
 – 5 is divisible by 24.
For n = 1,
P(1) : 2.7 + 3.5 – 5 = 24, which is divisible by 24.
Assume that P(k) is true,
i.e. 2.7
k
 + 3.5
k
 – 5 = 24q, where q ? N ... (i)
Now, we wish to prove that P(k + 1) is true whenever P(k) is true, i.e.
2.7
k + 1
 + 3.5
k + 1
 – 5 is divisible by 24.
We have,
2.7
k + 1
 + 3.5
k + 1
 – 5 = 2.7
k
 
.
 7
1
 + 3.5
k
 
.
 5
1
 – 5
= 7[2.7
k
 + 3.5
k
 – 5 – 3.5
k
 + 5] + 3.5
k
 
.
 5 – 5
= 7[24q – 3.5
k
 + 5] + 15.5
k
 – 5
= (7 × 24q) – 21.5
k
 + 35 + 15.5
k
 – 5
= (7 × 24q) – 6.5
k
 + 30 = (7× 24q) – 6(5
k
 – 5)
= (7 × 24q) – 6(4p)  [ (5
k
 – 5) is a multiple of 4]
= (7 × 24q) – 24p = 24(7q – p)
= 24 × r; r = 7q – p, is some natural number  ... (ii)
Thus, 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.
17. (b) Let P(n) : 
= . For n = 1,
L.H.S. = 
and   R.H.S. = 
?   P(1) is true.
Let P(k) is true, then
P(k) : 
= ... (i)
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? P.M.I cannot be applied
Let S(k) is true, i.e.
5. (b) Check through option, the condition  is satisfied for
n > 2
6. (a) Putting n = 2 in 3
2n
 –2n + 1 then,
 which is divisibleby 2.
7. (a) The product of two consecutive numbers is always even.
8. (b) a
1
 = < 7. Let a
m
 < 7
Then a
m
 
+ 1
 =   a
2
? m + 1
= 7 + a
m
 < 7 + 7 < 14.
 a
m + 1
 < < 7;  So by the principle of mathematical
induction a
n
 < 7 n. 
9. (c) Check through option, the condition is true when .
10. (d) Let P(n) : 
For n = 2,
P(2) :  ? 
which is true.
Let for n = m = 2, P(m) is true.
i.e.
Now,  = 
< 
= 
= 
< 
Hence, for n = 2, P(n) is true.
11. (a) x
2n–1
 + y
2n–1
 is always contain equal odd power. So it is always
divisible by x + y .
12. (a) It can be proved with the help of mathematical induction that  <
a(n) = n.
? < a(200) ? a(200) > 100 and
a(100) = 100.
13. (d) Let the given statement be P (n), then
P (1) ? 2
1
 > 1
2
 which is true
P (2) ? 2
2
 > 2
2
 which is false
P (3) ? 2
3
 > 3
2
 which is false
P (4) ? 2
4
 > 4
2
 which is false
P (5) ? 2
5
 > 5
2
 which is true
P (6) ? 2
6
 > 6
2
 which is true
?   P (n) is true when n = 5
14. (a) Let P (n) = n (n
2
 – 1) then
P(1) = 1 (0) = 0 which is divisible by every n ? N
P (3) = 3 (8) = 24 which is divisible by 24 and 8
P (5) = 5 (24) = 120 which is divisible by 24 and 8
Hence P (n) is divisible by 24.
15. (c) Let n = 1, then option (a), (b) and (d) eliminated. Only option (c)
satisfied.
16. (a) P(n) : 2.7
n
 + 3.5
n
 – 5 is divisible by 24.
For n = 1,
P(1) : 2.7 + 3.5 – 5 = 24, which is divisible by 24.
Assume that P(k) is true,
i.e. 2.7
k
 + 3.5
k
 – 5 = 24q, where q ? N ... (i)
Now, we wish to prove that P(k + 1) is true whenever P(k) is true, i.e.
2.7
k + 1
 + 3.5
k + 1
 – 5 is divisible by 24.
We have,
2.7
k + 1
 + 3.5
k + 1
 – 5 = 2.7
k
 
.
 7
1
 + 3.5
k
 
.
 5
1
 – 5
= 7[2.7
k
 + 3.5
k
 – 5 – 3.5
k
 + 5] + 3.5
k
 
.
 5 – 5
= 7[24q – 3.5
k
 + 5] + 15.5
k
 – 5
= (7 × 24q) – 21.5
k
 + 35 + 15.5
k
 – 5
= (7 × 24q) – 6.5
k
 + 30 = (7× 24q) – 6(5
k
 – 5)
= (7 × 24q) – 6(4p)  [ (5
k
 – 5) is a multiple of 4]
= (7 × 24q) – 24p = 24(7q – p)
= 24 × r; r = 7q – p, is some natural number  ... (ii)
Thus, 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.
17. (b) Let P(n) : 
= . For n = 1,
L.H.S. = 
and   R.H.S. = 
?   P(1) is true.
Let P(k) is true, then
P(k) : 
= ... (i)
For n = k + 1,
P(k + 1) : 
+ 
= 
L.H.S. = 
+ 
= 
[from (i)]
= = R.H.S.
Hence, P(k + 1) is true.
Hence, by principle of mathematical induction for all n ? N, P(n) is
true.
18. (c) Let P(n) : 7
n
 – 3
n
 is divisible by 4. For n = 1,
P(1) : 7
1
 – 3
1
 = 4, which is divisible by 4. Thus, P(n) is true for n = 1.
Let P(k) be true for some natural number k,
i.e.  P(k) : 7
k
 – 3
k
 is divisible by 4.
We can write 7
k
 – 3
k
 = 4d, where d ? N ... (i)
Now, we wish to prove that P(k + 1) is true whenever P(k) is true, i.e. 7
k
+ 1
 – 3
k + 1
 is divisible by 4.
Now, 7
(k + 1)
 – 3
(k + 1)
 = 7
(k + 1)
 – 7.3
k
 + 7.3
k
 – 3
(k + 1)
= 7(7
k
 – 3
k
) + (7 – 3)3
k
 = 7(4d) + 4.3
k    
[using (i)]
= 4(7d + 3
k
), which is divisible by 4.
Thus, P(k + 1) is true whenever P(k) is true. Therefore, by the principle
of mathematical induction the statement is true for every positive
integer n.