Question

Number of squares of any size in a rectangle of size \[p\times n\left( p < n \right)\] is given as
\[\sum\limits_{r=1}^{p}{\left( n+1-r \right)\left( p+1-r \right)}\]
(a) True
(b) False

Answer 1

Hint: We solve this problem by using all the condition that the minimum side length of the square as 1 unit to maximum side length of the square as \[p\] units because \[p < n\]
We find the number of squares in each condition for 3 to 4 types that is we find the number of squares considering the side length from 1 unit to 3 units and the final side length as \[p\] units so that we can generalize the equation in the summation form which is the required formula.
We use the condition that if there are \[x\] squares horizontally and \[y\] squares vertically then the total number of squares is given as \[x\times y\]

Complete step by step answer:
Let us consider the rectangle of size \[p\times n\left( p < n \right)\] such that it is divided into lines of 1 unit vertically and horizontally as shown below

Now, let us find the number of squares of minimum side length that is 1 unit.
Let us assume that the number of squares of length 1 unit as \[{{S}_{1}}\]
Here, we can see that there are a total of \[n\] squares horizontally and \[p\] squares vertically.
We know that the condition that if there are \[x\] squares horizontally and \[y\] squares vertically then the total number of squares is given as \[x\times y\]
By using this condition we get the number of squares of 1 unit length as
\[\Rightarrow {{S}_{1}}=n\times p\]
Now, let us find the number of squares of side length that is 2 units.
Let us assume that the number of squares of length 2 units as \[{{S}_{2}}\]
Here, we can see that there are a total of \[n-1\] squares horizontally and \[p-1\] squares vertically.
We know that the condition that if there are \[x\] squares horizontally and \[y\] squares vertically then the total number of squares is given as \[x\times y\]
By using this condition we get the number of squares of 2 unit length as
\[\Rightarrow {{S}_{2}}=\left( n-1 \right)\times \left( p-1 \right)\]
Now, let us find the number of squares of side length that is 3 units.
Let us assume that the number of squares of length 3 units as \[{{S}_{3}}\]
Here, we can see that there are a total of \[n-2\] squares horizontally and \[p-2\] squares vertically.
We know that the condition that if there are \[x\] squares horizontally and \[y\] squares vertically then the total number of squares is given as \[x\times y\]
By using this condition we get the number of squares of 2 unit length as
\[\Rightarrow {{S}_{3}}=\left( n-2 \right)\times \left( p-2 \right)\]
Now let us find the number of squares of maximum side length of \[p\] unit length because \[\left( p < n \right)\]
Let us assume that the number of squares of length \[p\] units as \[{{S}_{p}}\]
Here, we can see that there are total of \[\left( n-p+1 \right)\] squares horizontally and \[\left( p-p+1 \right)=1\] square vertically.
By using the total number of squares formula we get
\[\Rightarrow {{S}_{p}}=\left( n-p+1 \right)\times 1\]

Now, let us assume that the total number of squares as \[S\] then we get
\[\Rightarrow S={{S}_{1}}+{{S}_{2}}+{{S}_{3}}+.....+{{S}_{p}}\]
Now, by substituting the required values in above equation we get
\[\Rightarrow S=\left( n\times p \right)+\left( \left( n-1 \right)\left( p-1 \right) \right)+\left( \left( n-2 \right)\left( p-2 \right) \right)+......+\left( \left( n-p+1 \right)\left( 1 \right) \right)\]
Now, let us represent each term with respect to their side length then we get
\[\Rightarrow S=\left[ \left( n-1+1 \right)\left( p-1+1 \right) \right]+\left[ \left( n-2+1 \right)\left( p-2+1 \right) \right]+......+\left[ \left( n-p+1 \right)\left( p-p+1 \right) \right]\]
Now, let us convert the above equation into summation form taking the side length as variable then we get
\[\Rightarrow S=\sum\limits_{r=1}^{p}{\left( n-r+1 \right)\left( p-r+1 \right)}\]
Therefore we can conclude that the total number of squares as \[\sum\limits_{r=1}^{p}{\left( n-r+1 \right)\left( p-r+1 \right)}\]
So, option (a) is the correct answer.

Note:
Students may make mistakes in taking the number of squares in the last condition that is the side length of \[p\] units.
We have that there are a total of \[\left( n-p+1 \right)\] squares horizontally and \[\left( p-p+1 \right)=1\] square vertically.
But students may make mistakes and assume that there is a total of \[\left( n-p \right)\] squares horizontally and \[1\] square vertically.
This gives the wrong answer because we have the sequence as shown below.
Here we can see that for the number of squares horizontally of side length of 1 unit, 2units, and 3 units we have the number of squares in terms of side length as
For 1 unit length we have \[n=n+1-1\]
For 2 units length we have \[n-1=n-2+1\]
For 3 unit length we have \[n-2=n-3+1\]
Similarly, in this formation for \[p\] unit length we have \[n-p+1\]