Statement - 1: For every natural number n ≥ 2, 1 √1+ 1 √2+…+ 1 √n>√n. Statement -2 : For every natural number n ≥ 2, √n(n+1) < n+1.

$$\begin{aligned} & P(n)=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\ldots+\frac{1}{\sqrt{n}} \\ & P(2)=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}>\sqrt{2} \end{aligned}$$​P(n)=1​1​+2​1​+…+n​1​P(2)=1​1​+2​1​>2​​ Let us assume that $P(k)=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\ldots+\frac{1}{\sqrt{k}}>\sqrt{k}$P(k)=1​1​+2​1​+…+k​1​>k​ is true $\therefore P(k+1)=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\ldots+\frac{1}{\sqrt{k}}+\frac{1}{\sqrt{k+1}}>\sqrt{k+1}$∴P(k+1)=1​1​+2​1​+…+k​1​+k+1​1​>k+1​ has to be true. L.H.S. $>\sqrt{\mathrm{k}}+\frac{1}{\sqrt{\mathrm{k}+1}}=\frac{\sqrt{\mathrm{k}(\mathrm{k}+1)}+1}{\sqrt{\mathrm{k}+1}}$>k​+k+1​1​=k+1​k(k+1)​+1​ Since $\sqrt{k(k+1)}>k \quad(\forall k \geq 0)$k(k+1)​>k(∀k≥0) $$\therefore \frac{\sqrt{\mathrm{k}(\mathrm{k}+1)}+1}{\sqrt{\mathrm{k}+1}}>\frac{\mathrm{k}+1}{\sqrt{\mathrm{k}+1}}=\sqrt{\mathrm{k}+1}$$∴k+1​k(k+1)​+1​>k+1​k+1​=k+1​ Let $P(n)=\sqrt{n(n+1)} < n+1$P(n)=n(n+1)​<n+1 Statement $-1$−1 is correct. $$P(2)=\sqrt{2 \times 3} < 3$$P(2)=2×3​<3 If $P(k)=\sqrt{k(k+1)} < (k+1)$P(k)=k(k+1)​<(k+1) is true Now $P(k+1)=\sqrt{(k+1)(k+2)} < k+2$P(k+1)=(k+1)(k+2)​<k+2 has to be true Since $(k+1) < k+2$(k+1)<k+2 $$\therefore \sqrt{(\mathrm{k}+1)(\mathrm{k}+2)} < (\mathrm{k}+2)$$∴(k+1)(k+2)​<(k+2) Hence Statement $-2$−2 is not a correct explanation of Statement $-1$−1.