LaTeX in Markdown Test
Post created: September 25, 2024Question
One can easily prove (by M.I.) that \(f(n) = \frac{n(n+1)(2n+1)}{6}, \forall n \geq 0\)
Proof
Let $f(n)$ be $0^2+1^2+2^2+3^2+\ldots+n^2=\frac{n(n+1)(2n+1)}{6}$ , for $\forall n \geq 0$,
\[\begin{align*} \text{For } n = 0, \ \text{L.H.S.} = 0 \\ \text{R.H.S.} = 0 \end{align*}\] \[\because \text{L.H.S.} = \text{R.H.S.}\] \[\therefore f(0) \text{ is true.}\]Assume $S(n)$ is true for some $n=k$ where $k \geq 0$, i.e. $0^2+1^2+2^2+3^2+\ldots+k^2=\frac{k(k+1)(2k+1)}{6}$
\[\begin{align*} \text{For $n=k+1$ , L.H.S.} &= 0^2+1^2+2^2+3^2+\ldots+k^2+(k+1)^2 \\ &= \frac{k(k+1)(2k+1)}{6}+(k+1)^2 \text{ (By induction assumption)} \\ &= \frac{k(k+1)(2k+1)+6(k+1)^2}{6} \\ &= \frac{(k+1)(k(2k+1)+6(k+1))}{6} \\ &= \frac{(k+1)(2k^2+7k+6)}{6} \\ &= \frac{(k+1)(k+2)(2k+3)}{6} \\ &= \frac{(k+1)((k+1)+1)(2(k+1)+1)}{6} \\ &= \text{R.H.S.} \end{align*}\] \[\because \text{L.H.S.} = \text{R.H.S.}\] \[\therefore \text{By the principle of mathematical induction, } f(n) \text{ is true for all } n \geq 0.\]