Gujarat BoardEnglish MediumSTD 11 ScienceMATHSPRINCIPLE OF MATHEMATICAL INDUCTION1 Mark
MCQ
$\mathrm{n}^2+3 \mathrm{n}$ is always divisible by which number, provided n is an integer?
✓
2
B
3
C
4
D
5
✓
Answer
Correct option: A.
2
2
Solution:
$P(n)=n^2+3 n$
$P(1)=1+3$
$P(1)=4$
Let's assume that $P(k)$ is true and divisible by 4 . Therefore, $P(k)=k^2+3 k$ can be written as $4 c$.
We need to check if $P(k+1)$ is divisible by 4
$\mathrm{P}(\mathrm{k}+1)=(\mathrm{k}+1)^2+3(\mathrm{k}+1)$
$\mathrm{P}(\mathrm{k}+1)=\mathrm{k}^2+1+2 \mathrm{k}+3 \mathrm{k}+3$
$\mathrm{P}(\mathrm{k}+1)=\mathrm{k}^2+5 \mathrm{k}+4$
$\mathrm{P}(\mathrm{k}+1)=\left(\mathrm{k}^2+3 \mathrm{k}\right)+2 \mathrm{k}+4$
$\mathrm{P}(\mathrm{k}+1)=4 \mathrm{c}+2 \mathrm{k}+4$
$\mathrm{P}(\mathrm{k}+1)=4 \mathrm{c}+2(\mathrm{k}+2)$
Clearly the second part of the equation is not divisible by 4 . However $P(k)=4 c$ is divisible by 2 and $P(k+1)$ is also divisible by 2 . Therefore, 2 divides $P(n)$
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