JEE Main 2021 — Mathematics Calculus

We know that $r\leq [r]<r+1$r≤[r]<r+1

$$2r\leq [2r]<2r+1$$2r≤[2r]<2r+1

$$3r\leq [3r]<3r+1$$3r≤[3r]<3r+1

$$\vdots \vdots \vdots$$⋮⋮⋮

$$nr\leq [nr]<nr+1$$nr≤[nr]<nr+1

On adding all the above inequalities, we get

$$r+2r+\ldots +nr\leq [r]+[2r]+...+[nr]<r+2r+...nr+n$$r+2r+…+nr≤[r]+[2r]+...+[nr]<r+2r+...nr+n

Using $1+2+3+...+n=\frac{n(n+1)}{2},n\in N,$1+2+3+...+n=2n(n+1)​,n∈N,

$$\frac{n(n+1)}{2}\cdot r\leq [r]+[2r]+...+[nr]<\frac{n(n+1)}{2}\cdot r+n$$2n(n+1)​⋅r≤[r]+[2r]+...+[nr]<2n(n+1)​⋅r+n

$$\Rightarrow \frac{\frac{n(n+1)}{2}\cdot r}{{n}^{2}}\leq \frac{[r]+[2r]+...+[nr]}{{n}^{2}}<\frac{\frac{n(n+1)}{2}\cdot r+n}{{n}^{2}}$$⇒n22n(n+1)​⋅r​≤n2[r]+[2r]+...+[nr]​<n22n(n+1)​⋅r+n​

Now, $\underset{n\rightarrow \infty }{\mathrm{lim}}\frac{n(n+1)\cdot r}{2\cdot {n}^{2}}=\underset{n\rightarrow \infty }{\mathrm{lim}}\frac{{n}^{2}(1+\frac{1}{n})\cdot r}{2\cdot {n}^{2}}=\frac{r}{2}$n→∞lim​2⋅n2n(n+1)⋅r​=n→∞lim​2⋅n2n2(1+n1​)⋅r​=2r​

and $\underset{n\rightarrow \infty }{\mathrm{lim}}\frac{\frac{n(n+1)}{2}\cdot r+n}{{n}^{2}}=\underset{n\rightarrow \infty }{\mathrm{lim}}\frac{{n}^{2}{(1+\frac{1}{n})\cdot r+\frac{1}{n}}}{2\cdot {n}^{2}}=\frac{r}{2}$n→∞lim​n22n(n+1)​⋅r+n​=n→∞lim​2⋅n2n2(1+n1​)⋅r+n1​​=2r​

So, by Sandwich Theorem, we can conclude that

$$\underset{n\rightarrow \infty }{\mathrm{lim}}\frac{[r]+[2r]+...+[nr]}{{n}^{2}}=\frac{r}{2}.$$n→∞lim​n2[r]+[2r]+...+[nr]​=2r​.