CBSE BoardEnglish MediumSTD 11 ScienceMathsPrinciple of Mathematical Induction1 Mark
MCQ
For any natural number $n, 2^{2n} - 1$ is divisible by:
A
$2$
✓
$3$
C
$4$
D
$5$
✓
Answer
Correct option: B.
$3$
Let $P(n)=2^{2 n}-1$
Substituting $n = 1, 2, 3,….$
$P(1)=2^{2(1)}-1=4-1=3$
This is divisible by $3.$
$P(2)=2^{2(2)}-1=16-1=15$
This is divisible by $3.$
$P(3)=2^{2(3)}-1=256-1=255$
This is also divisible by $3.$
Assume that $P(n)$ is true for some natural number $k,$
i.e., $P(k): 2^{2k}- 1$ is divisible by $3,$
i.e., $2^{2k}- 1 = 3q$,
where $q \in N$
Now,
$P(k+1): 2^{2(k+1)}-1$
$=2^{2 k+2}-1$
$=2^{2 k} \times 2^2-1$
$=2^{2 k} \times 4-1$
$=3.2^{2 k}+\left(2^{2 k}-1\right)$
$=3.2^{2 k}+3 q$
$=3\left(2^{2 k}+q\right)=3m,$
where $m \in N$
Thus $P(k + 1)$ is true, whenever $P(k)$ is true.
Therefore, for any natural number $n, 2^{2n-1}$ is divisible by $3$
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