Q.(a) By Mathematical Induction, prove that $1^2+2^2+3^2+\dots+n^2=\dfrac{n(n+1)(2n+1)}{6}$, for all $n\in N$.

(a) The sum-of-squares formula is proved by induction. (b) $\dfrac{dy}{dx}=\dfrac{\sin^2(a+y)}{\sin a}$.

(a) Induction: $1^2+\dots+n^2=\dfrac{n(n+1)(2n+1)}{6}$

Base $n=1$: LHS $=1$, RHS $=\dfrac{1\cdot2\cdot3}{6}=1$ ✓.

Assume for $k$: $\sum_{r=1}^k r^2=\dfrac{k(k+1)(2k+1)}{6}.$ Add $(k+1)^2$:

$\frac{k(k+1)(2k+1)}{6}+(k+1)^2=(k+1)\frac{k(2k+1)+6(k+1)}{6}=(k+1)\frac{2k^2+7k+6}{6}=\frac{(k+1)(k+2)(2k+3)}{6},$

which is the formula for $k+1$. By induction it holds for all $n\in\mathbb N$.

(b) If $\sin y=x\sin(a+y)$

$x=\dfrac{\sin y}{\sin(a+y)}.$ Differentiate w.r.t. $y$ (quotient rule): …