12th Grade > Mathematics
PRINCIPLE OF MATHEMATICAL INDUCTION MCQs
Total Questions : 15
| Page 2 of 2 pages
Answer: Option A. -> 25
:
A
Putting n = 1 in 72n+23n−3.3n−1
=50, divisible by 25
:
A
Putting n = 1 in 72n+23n−3.3n−1
=50, divisible by 25
Answer: Option A. -> x + y
:
A
x2n−1+y2n−1 is always contain equal odd power.
So it is always divisible by x + y.
:
A
x2n−1+y2n−1 is always contain equal odd power.
So it is always divisible by x + y.
Answer: Option C. -> >n33
:
C
Let n=1
n32=12;n33=13
Let n=2,⇒12+22=5
n32=4;n33=83
Letn=3,⇒12+22+32=14
n32=272;n33=9
Letn=4,⇒12+22+32+42=30
n32=32;n33=643
12+22+32+⋯+n2>n33
P(n):12+22+32+⋯+n2>n33
P(1) is true
Let P(k) be true.
⇒12+22+32+⋯+k2>k33
12+22+32+⋯+k2+(k+1)2>k33+k2+2k+1
=k3+3k2+6k+33
=k3+3k2+3k+1+3k+23
=(k+1)33+k+23
12+22+32+⋯+k2+(k+1)2>(k+1)33
P(k+1) is true.
P(n) is true ∀n∈N
:
C
Let n=1
n32=12;n33=13
Let n=2,⇒12+22=5
n32=4;n33=83
Letn=3,⇒12+22+32=14
n32=272;n33=9
Letn=4,⇒12+22+32+42=30
n32=32;n33=643
12+22+32+⋯+n2>n33
P(n):12+22+32+⋯+n2>n33
P(1) is true
Let P(k) be true.
⇒12+22+32+⋯+k2>k33
12+22+32+⋯+k2+(k+1)2>k33+k2+2k+1
=k3+3k2+6k+33
=k3+3k2+3k+1+3k+23
=(k+1)33+k+23
12+22+32+⋯+k2+(k+1)2>(k+1)33
P(k+1) is true.
P(n) is true ∀n∈N
Answer: Option A. -> 2
:
A
Putting n = 2 in 32n - 2n + 1 then,
32×2 - 2×2+1 = 81 - 4 + 1 = 78, which is divisible
by 2.
:
A
Putting n = 2 in 32n - 2n + 1 then,
32×2 - 2×2+1 = 81 - 4 + 1 = 78, which is divisible
by 2.
Answer: Option B. -> n≥ 4
:
B
Check through option, the condition 2n < n! is
true when n≥ 4.
:
B
Check through option, the condition 2n < n! is
true when n≥ 4.
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