12th Grade > Mathematics
PRINCIPLE OF MATHEMATICAL INDUCTION MCQs
Total Questions : 15
| Page 1 of 2 pages
Answer: Option C. -> n≥ 4
:
C
Check through option, the condition 3n > n3 is
true when n≥ 4.
:
C
Check through option, the condition 3n > n3 is
true when n≥ 4.
Answer: Option B. -> n
:
B
Let n = 1 then option (a) and (d) is eliminated.
Equality can't be attained for any value of n so,
option (b) satisfied.
:
B
Let n = 1 then option (a) and (d) is eliminated.
Equality can't be attained for any value of n so,
option (b) satisfied.
Answer: Option C. -> n4 < 10n
:
C
Let n = 1, then option (a), (b) and (d)
eliminated. Only option (c) satisfied.
:
C
Let n = 1, then option (a), (b) and (d)
eliminated. Only option (c) satisfied.
Answer: Option D. -> n≥ 3
:
D
Check through option, condition (n!)2 > nn is
true when n≥ 3.
:
D
Check through option, condition (n!)2 > nn is
true when n≥ 3.
Answer: Option C. -> 133
:
C
Putting n = 1 in 11n+2+122n+1
We get, 111+2+122×1+1 = 113+123 = 3059, which
is divisible by 133.
:
C
Putting n = 1 in 11n+2+122n+1
We get, 111+2+122×1+1 = 113+123 = 3059, which
is divisible by 133.
Answer: Option B. -> 6
:
B
n(n2−1) = (n - 1)(n)(n + 1)
It is product of three consecutive natural
numbers, so according to Langrange's theorem
it is divisible by 3 ! i.e., 6.
:
B
n(n2−1) = (n - 1)(n)(n + 1)
It is product of three consecutive natural
numbers, so according to Langrange's theorem
it is divisible by 3 ! i.e., 6.
Answer: Option B. -> n≥ 5
:
B
Check through option, the condition
10n−2 > 81n is satisfied if n≥ 5.
:
B
Check through option, the condition
10n−2 > 81n is satisfied if n≥ 5.
Answer: Option B. -> n > 2
:
B
Check through option, the condition
2n(n-1)!<nn is satisfied for n > 2.
:
B
Check through option, the condition
2n(n-1)!<nn is satisfied for n > 2.
Answer: Option D. -> None of these
:
D
P(n) = n2 + n. It is always odd (statement) but
square of any odd number is always odd and
also, sum of odd number is always even. So
for no any 'n' for which this statement is true.
:
D
P(n) = n2 + n. It is always odd (statement) but
square of any odd number is always odd and
also, sum of odd number is always even. So
for no any 'n' for which this statement is true.
Answer: Option B. -> n≥ 1
:
B
Check through option, the condition
(n+12)n≥ n ! is true for n≥ 1.
:
B
Check through option, the condition
(n+12)n≥ n ! is true for n≥ 1.
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