Mathematics - Quadratic Equation Question with Solution | TestHub

$$\begin{array}{rl}& P(n)=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\dots +\frac{1}{\sqrt{n}}\\ & P(2)=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}>\sqrt{2}\end{array}$$ Let us assume that $P(k)=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\dots +\frac{1}{\sqrt{k}}>\sqrt{k}$ is true $\therefore P(k+1)=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\dots +\frac{1}{\sqrt{k}}+\frac{1}{\sqrt{k+1}}>\sqrt{k+1}$ has to be true. L.H.S. $>\sqrt{\mathrm{k}}+\frac{1}{\sqrt{\mathrm{k}+1}}=\frac{\sqrt{\mathrm{k}(\mathrm{k}+1)}+1}{\sqrt{\mathrm{k}+1}}$ Since $\sqrt{k(k+1)}>k(\forall k\ge 0)$ $$\therefore \frac{\sqrt{\mathrm{k}(\mathrm{k}+1)}+1}{\sqrt{\mathrm{k}+1}}>\frac{\mathrm{k}+1}{\sqrt{\mathrm{k}+1}}=\sqrt{\mathrm{k}+1}$$ Let $P(n)=\sqrt{n(n+1)}<n+1$ Statement $-1$ is correct. $$P(2)=\sqrt{2\times 3}<3$$ If $P(k)=\sqrt{k(k+1)}<(k+1)$ is true Now $P(k+1)=\sqrt{(k+1)(k+2)}<k+2$ has to be true Since $(k+1)<k+2$ $$\therefore \sqrt{(\mathrm{k}+1)(\mathrm{k}+2)}<(\mathrm{k}+2)$$ Hence Statement $-2$ is not a correct explanation of Statement $-1$.