(a) By the principle of Mathematical Induction, prove that for $n \ge 1$, $1\cdot2 + 2\cdot3 + 3\cdot4 + \ldots n(n+1) = \dfrac{n(n+1)(n+2)}{3}$. **OR** (b) In a $\triangle ABC$, prove that $\dfrac{a\sin(B-C)}{b^2-c^2} = \dfrac{b\sin(C-A)}{c^2-a^2} = \dfrac{c\sin(A-B)}{a^2-b^2}$.

(a) By the principle of Mathematical Induction, prove that for n ≥ 1, 1·2 + 2·3 + 3·4 + … n(n+1) = (n(n+1)(n+2

Question 113 of 134

n \ge 1

(b) In a

\triangle ABC

, prove that

\dfrac{a\sin(B-C)}{b^2-c^2} = \dfrac{b\sin(C-A)}{c^2-a^2} = \dfrac{c\sin(A-B)}{a^2-b^2}

Puducherry TnboardTamil Nadu HSC First Year (DGE) Board 2020Subjective· 5mImportance★★★★★

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Verified for n=1, then shown that if the formula holds for n=k it must hold for n=k+1, so by mathematical induction it holds for every $n \ge 1$ .

Let $P(n)$ : $1\cdot2+2\cdot3+3\cdot4+\ldots+n(n+1) = \dfrac{n(n+1)(n+2)}{3}$ .

Base case (n=1): LHS $=1\cdot2=2$ . RHS $=\dfrac{1\cdot2\cdot3}{3}=2$ . So $P(1)$ is true.

Inductive hypothesis: Assume $P(k)$ is true for some $k\ge1$ :

$1\cdot2+2\cdot3+\ldots+k(k+1) = \dfrac{k(k+1)(k+2)}{3}$

Inductive step: Show $P(k+1)$ holds. Add the next term $(k+1)(k+2)$ to both sides:

$1\cdot2+\ldots+k(k+1)+(k+1)(k+2) = \dfrac{k(k+1)(k+2)}{3}+(k+1)(k+2)$

Factor out $(k+1)(k+2)$ on the right: …