Q.(a) By mathematical Induction, prove that $1^3 + 2^3 + 3^3 + \ldots + n^3 = \dfrac{n^2(n+1)^2}{4}$ for all $n \in N$.

(a) By induction $1^3+\dots+n^3 = \dfrac{n^2(n+1)^2}{4}$: base holds, and the inductive step gives $\dfrac{(k+1)^2(k+2)^2}{4}$. (b) At $x=2$, LHL $=0 e$ RHL $=4 = f(2)$, so $f$ is discontinuous.

Part (a): Prove $1^3+2^3+\dots+n^3 = \dfrac{n^2(n+1)^2}{4}$ for all $n\in N$.

Base case $n=1$: LHS $= 1^3 = 1$; RHS $= \dfrac{1^2\cdot 2^2}{4} = \dfrac{4}{4} = 1$. True.

Inductive hypothesis: assume it holds for $n=k$:

$1^3+2^3+\dots+k^3 = \frac{k^2(k+1)^2}{4}.$

Inductive step ($n=k+1$):

$1^3+\dots+k^3+(k+1)^3 = \frac{k^2(k+1)^2}{4} + (k+1)^3 = (k+1)^2\left[\frac{k^2}{4} + (k+1)\right].$

$= (k+1)^2\cdot\frac{k^2 + 4k + 4}{4} = (k+1)^2\cdot\frac{(k+2)^2}{4} = \frac{(k+1)^2\big((k+1)+1\big)^2}{4}.$

This is the formula for $n=k+1$. Hence, by the principle of mathematical induction, the result holds for all $n\in N$.

Part (b): Continuity of $f(x) = \begin{cases}2-x, & x<2\\ 2+x, & x\ge 2\end{cases}$ at $x=2$.

Left-hand limit: $\displaystyle\lim_{x\to 2^-}(2-x) = 2-2 = 0.$ …