JEE Main 2014 — Mathematics Algebra

Given $f(n)=[\frac{1}{3}+\frac{3n}{100}]n=[\frac{9n+100}{300}]n$f(n)=[31​+1003n​]n=[3009n+100​]n

If $\frac{9n+100}{300}<1$3009n+100​<1

$$\Rightarrow 9n+100<300$$⇒9n+100<300

$$\Rightarrow n<\frac{200}{9}$$⇒n<9200​

$$\Rightarrow n=22$$⇒n=22

We know that for a number $0\leqslant x<1,[x]=0$0⩽x<1,[x]=0

$$\Rightarrow f(n)=0,\forall 1\leqslant n\leqslant 22$$⇒f(n)=0,∀1⩽n⩽22

Again, if $\frac{9n+100}{300}<2$3009n+100​<2

$$\Rightarrow 9n+100<600$$⇒9n+100<600

$$\Rightarrow n=\frac{500}{9}$$⇒n=9500​

$$\Rightarrow n=55$$⇒n=55

We know that for a number $1\leqslant x<2,[x]=1$1⩽x<2,[x]=1

$$\Rightarrow f(n)=1,23\leqslant n\leqslant 55$$⇒f(n)=1,23⩽n⩽55

And, $f(n)=2$f(n)=2 if $n=56$n=56

Thus, $\displaystyle \sum _{n=1}^{56}f(n)=\sum _{n=1}^{22}0\times n+\sum _{n=23}^{55}1\times n+2\times 56$n=1∑56​f(n)=n=1∑22​0×n+n=23∑55​1×n+2×56

$$=0+(23+24+...+55)+56\times 2$$=0+(23+24+...+55)+56×2

$$=\frac{33}{2}(23+55)+112$$=233​(23+55)+112

$$=33\times 39+112$$=33×39+112

$$=1399.$$=1399.