Question

A square of $n$ units by $n$ units is divided into $n^2$ squares each of area $1sq.unit$. Let $\frac{n^2(n+k)}{m}$ be the number of ways in which $4$ points (out of $(n+1{)}^2$ vertices of the squares) can be chosen so that they form the vertices of a square. Find $k+m$ ?

Answer 1

No. of squares of area $n^2$ square units $=1^2$
No. of squares of area $(n-1{)}^2$ square units $=2^2$
No. of squares of area $(n-2{)}^2$ square units $=3^2$
$.............................................$
No. of squares of area $1^2$ square units $=n^2$
Adding gives $N_1=1^2+2^2+3^2+...+n^2$
$=\frac{n(n+1)(2n+1)}{6}$
When $n$ is even:
No. of squares of area $\frac{n^2}{2}$ square units $=1^2$
No. of squares of area $\frac{(n-2{)}^2}{2}$ square units $=3^2$
$.......................................$
No. of squares of area $\frac{2^2}{2}$ square units $=(n-1{)}^2$
Adding gives $N_2=1^2+3^2+5^2+...+(n-1{)}^2$
$=\frac{n(n-1)(n+1)}{6}$
When $n$ is odd:
No. of squares of area $\frac{(n-1{)}^2}{2}$ square units $=2^2$
No. of squares of area $\frac{(n-3{)}^2}{2}$ square units $=4^2$
No. of squares of area $\frac{(n-5{)}^2}{2}$ square units $=6^2$
$.......................................$
No. of squares of area $\frac{2^2}{2}$ square units $=(n-1{)}^2$
Adding gives $N_2=2^2+4^2+6^2+...+(n-1{)}^2$
$=\frac{n(n-1)(n+1)}{6}$
$\therefore$ Total no. of squares formed which can be obtained by taking $4$ points out of $(n+1{)}^2$ points $=N_1+N_2$
$=\frac{n(n+1)(2n+1)}{6}+\frac{n(n-1)(n+1)}{6}$
$=\frac{n^2(n+1)}{2}$
$\Rightarrow k+m=1+2=3$