Q.Prove that the sum of the first $n$ odd natural numbers is $n^2$, i.e. $1 + 3 + 5 + \cdots + (2n-1) = n^2$, for all $n \ge 1$.

Base case $n=1$ holds ($1<9/8$); the inductive step shows the bound for $n=k+1$ equals the bound for $n=k$ plus exactly $(k+1)$, so the strict inequality is preserved for all $n$.

Note: This topic (Principle of Mathematical Induction) was a standalone NCERT chapter in earlier syllabi but has been removed from the current rationalised Class 11 syllabus — it is included here for completeness since it was asked in this year's paper.

Let $P(n): 1+2+3+\cdots+n < \dfrac18(2n+1)^2$.

Base case ($n=1$): LHS $=1$. RHS $=\dfrac18(3)^2=\dfrac98=1.125$. Since $1<1.125$, $P(1)$ is true.

Inductive step: Assume $P(k)$ is true: $1+2+\cdots+k < \dfrac18(2k+1)^2$.

We must show $P(k+1)$: $1+2+\cdots+k+(k+1) < \dfrac18(2k+3)^2$.

From the inductive hypothesis,

$1+2+\cdots+k+(k+1) < \dfrac18(2k+1)^2 + (k+1)$.

Now compare $\dfrac18(2k+1)^2+(k+1)$ with $\dfrac18(2k+3)^2$:

$(2k+3)^2 - (2k+1)^2 = [(2k+3)-(2k+1)][(2k+3)+(2k+1)] = 2(4k+4) = 8k+8$.

So $\dfrac18(2k+3)^2 - \dfrac18(2k+1)^2 = \dfrac{8k+8}{8} = k+1$.

That means $\dfrac18(2k+1)^2+(k+1) = \dfrac18(2k+3)^2$ exactly.

Therefore $1+2+\cdots+(k+1) < \dfrac18(2k+1)^2+(k+1) = \dfrac18(2k+3)^2$, i.e. $P(k+1)$ is true.

By the principle of mathematical induction, $P(n)$ is true for all $n\in\mathbb N$.

Proved by mathematical induction: $1^3+2^3+\cdots+n^3=\left(\dfrac{n(n+1)}{2}\right)^2$ for every natural number $n$.

Let $P(n): 1^3+2^3+\cdots+n^3=\left(\dfrac{n(n+1)}{2}\right)^2$.

Base case ($n=1$): LHS $=1^3=1$. RHS $=\left(\dfrac{1\cdot2}{2}\right)^2=1^2=1$. Since LHS = RHS, $P(1)$ is true.

Inductive step: Assume $P(k)$ is true, i.e.

$1^3+2^3+\cdots+k^3=\left(\frac{k(k+1)}{2}\right)^2$

We must show $P(k+1)$: $1^3+\cdots+k^3+(k+1)^3=\left(\dfrac{(k+1)(k+2)}{2}\right)^2$.

Adding $(k+1)^3$ to both sides of the assumption:

$1^3+\cdots+k^3+(k+1)^3 = \left(\frac{k(k+1)}{2}\right)^2+(k+1)^3$

Factor out $\dfrac{(k+1)^2}{4}$ from the right side:

$= \frac{(k+1)^2}{4}\Big[k^2+4(k+1)\Big] = \frac{(k+1)^2}{4}\big[k^2+4k+4\big] = \frac{(k+1)^2(k+2)^2}{4} = \left(\frac{(k+1)(k+2)}{2}\right)^2$

This is exactly $P(k+1)$. Hence by the principle of mathematical induction, $P(n)$ is true for all $n\in N$.