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Question: If \[{\Delta _r} = \left| {\begin{array}{*{20}{c}} r&{2r - 1}&{3r + 2} \\\ {\dfrac{n}{2}}&{...

If {\Delta _r} = \left| {\begin{array}{*{20}{c}} r&{2r - 1}&{3r + 2} \\\ {\dfrac{n}{2}}&{n - 1}&a; \\\ {\dfrac{1}{2}n\left( {n - 1} \right)}&{{{\left( {n - 1} \right)}^2}}&{\dfrac{1}{2}\left( {n - 1} \right)\left( {3n + 4} \right)} \end{array}} \right|, then the value of∑r=1n−1Δr\sum\limits_{r = 1}^{n - 1} {\Delta r} :
A) Depends only on a
B) Depends only on n
C) Depends both on a and n
D) Is Independent of both a and n

Explanation

Solution

Here we will use the values of summation and then apply elementary row operations to evaluate the value of given determinant.
∑x=1n−1x=n(n−1)2\sum\limits_{x = 1}^{n - 1} x = \dfrac{{n\left( {n - 1} \right)}}{2}

Complete step-by-step answer:
The given expression is:-

r&{2r - 1}&{3r + 2} \\\ {\dfrac{n}{2}}&{n - 1}&a; \\\ {\dfrac{1}{2}n\left( {n - 1} \right)}&{{{\left( {n - 1} \right)}^2}}&{\dfrac{1}{2}\left( {n - 1} \right)\left( {3n + 4} \right)} \end{array}} \right|$$ We will first evaluate the value of $$\sum\limits_{r = 1}^{n - 1} r $$. Using the following formula:- $$\sum\limits_{x = 1}^{n - 1} x = \dfrac{{n\left( {n - 1} \right)}}{2}$$ We get:- $$ \Rightarrow \sum\limits_{r = 1}^{n - 1} r = \dfrac{{n\left( {n - 1} \right)}}{2}$$……………………….(1) Now we will evaluate the value of$$\sum\limits_{x = 1}^{n - 1} {2r - 1} $$. Splitting the summation we get:- $$ \Rightarrow \sum\limits_{x = 1}^{n - 1} {2r - 1} = \sum\limits_{x = 1}^{n - 1} {2r} - \sum\limits_{x = 1}^{n - 1} 1 $$ Solving it further we get:- $$ \Rightarrow \sum\limits_{x = 1}^{n - 1} {2r - 1} = 2\sum\limits_{x = 1}^{n - 1} r - \sum\limits_{x = 1}^{n - 1} 1 $$ Now we know that:- $$\sum\limits_{r = 1}^{n - 1} 1 = n - 1$$ Putting this value in above equation we get:- $$ \Rightarrow \sum\limits_{x = 1}^{n - 1} {2r - 1} = 2\sum\limits_{x = 1}^{n - 1} r - \left( {n - 1} \right)$$ Now putting the value from equation 1 we get:- $$ \Rightarrow \sum\limits_{x = 1}^{n - 1} {2r - 1} = 2\left[ {\dfrac{{n\left( {n - 1} \right)}}{2}} \right] - \left( {n - 1} \right)$$ Solving it further we get:- $$ \Rightarrow \sum\limits_{x = 1}^{n - 1} {2r - 1} = n\left( {n - 1} \right) - \left( {n - 1} \right) \\\ \Rightarrow \sum\limits_{x = 1}^{n - 1} {2r - 1} = {n^2} - n - n + 1 \\\ \Rightarrow \sum\limits_{x = 1}^{n - 1} {2r - 1} = {\left( {n - 1} \right)^2}.................\left( 2 \right) \\\

Now we will evaluate the value of ∑x=1n−13r+2\sum\limits_{x = 1}^{n - 1} {3r + 2}
Splitting the summation we get:-
∑x=1n−13r+2=∑r=1n−13r+∑r=1n−12\sum\limits_{x = 1}^{n - 1} {3r + 2} = \sum\limits_{r = 1}^{n - 1} {3r + \sum\limits_{r = 1}^{n - 1} 2 }
Solving it further we get:-
⇒∑x=1n−13r+2=3∑r=1n−1r+2∑r=1n−11\Rightarrow \sum\limits_{x = 1}^{n - 1} {3r + 2} = 3\sum\limits_{r = 1}^{n - 1} {r + 2\sum\limits_{r = 1}^{n - 1} 1 }
Now we know that:-
∑r=1n−11=n−1\sum\limits_{r = 1}^{n - 1} 1 = n - 1
Putting this value in above equation we get:-
⇒∑x=1n−13r+2=3∑r=1n−1r+2(n−1)\Rightarrow \sum\limits_{x = 1}^{n - 1} {3r + 2} = 3\sum\limits_{r = 1}^{n - 1} r + 2\left( {n - 1} \right)
Now putting the value from equation 1 we get:-
∑x=1n−13r+2=3[n(n−1)2]+2n−2\sum\limits_{x = 1}^{n - 1} {3r + 2} = 3\left[ {\dfrac{{n\left( {n - 1} \right)}}{2}} \right] + 2n - 2
Taking LCM we get:-

∑x=1n−13r+2=3n(n−1)+2(2n−2)2 ⇒∑x=1n−13r+2=3n2−3n+4n2−4n2  \sum\limits_{x = 1}^{n - 1} {3r + 2} = \dfrac{{3n\left( {n - 1} \right) + 2\left( {2n - 2} \right)}}{2} \\\ \Rightarrow \sum\limits_{x = 1}^{n - 1} {3r + 2} = \dfrac{{3{n^2} - 3n + 4{n^2} - 4n}}{2} \\\

Solving it further we get:-
∑x=1n−13r+2=3n(n−1)+2(2n−2)2\sum\limits_{x = 1}^{n - 1} {3r + 2} = \dfrac{{3n\left( {n - 1} \right) + 2\left( {2n - 2} \right)}}{2}
Taking 2 as common we get:-
∑x=1n−13r+2=3n(n−1)+2(2)(n−1)2\sum\limits_{x = 1}^{n - 1} {3r + 2} = \dfrac{{3n\left( {n - 1} \right) + 2\left( 2 \right)\left( {n - 1} \right)}}{2}
Solving it further we get:-

∑x=1n−13r+2=(n−1)[3n+2(2)]2 ⇒∑x=1n−13r+2=(n−1)[3n+4]2..........................(3)  \sum\limits_{x = 1}^{n - 1} {3r + 2} = \dfrac{{\left( {n - 1} \right)\left[ {3n + 2\left( 2 \right)} \right]}}{2} \\\ \Rightarrow \sum\limits_{x = 1}^{n - 1} {3r + 2} = \dfrac{{\left( {n - 1} \right)\left[ {3n + 4} \right]}}{2}..........................\left( 3 \right) \\\

Now we will evaluate the value of ∑r=1n−1Δr\sum\limits_{r = 1}^{n - 1} {\Delta r} .
Hence we get:-

{\sum\limits_{r = 1}^{n - 1} r }&{\sum\limits_{x = 1}^{n - 1} {2r - 1} }&{\sum\limits_{x = 1}^{n - 1} {3r + 2} } \\\ {\dfrac{n}{2}}&{n - 1}&a; \\\ {\dfrac{1}{2}n\left( {n - 1} \right)}&{{{\left( {n - 1} \right)}^2}}&{\dfrac{1}{2}\left( {n - 1} \right)\left( {3n + 4} \right)} \end{array}} \right|$$ Now putting the values from equation 1, 2 and 3 we get:- $$\sum\limits_{r = 1}^{n - 1} {{\Delta _r}} = \left| {\begin{array}{*{20}{c}} {\dfrac{{n\left( {n - 1} \right)}}{2}}&{{{\left( {n - 1} \right)}^2}}&{\dfrac{{\left( {n - 1} \right)\left[ {3n + 4} \right]}}{2}} \\\ {\dfrac{n}{2}}&{n - 1}&a; \\\ {\dfrac{1}{2}n\left( {n - 1} \right)}&{{{\left( {n - 1} \right)}^2}}&{\dfrac{1}{2}\left( {n - 1} \right)\left( {3n + 4} \right)} \end{array}} \right|$$ Now since we know that if two rows of a determinant are same then its value is zero Therefore, $$\sum\limits_{r = 1}^{n - 1} {{\Delta _r}} = 0$$ Hence it is independent of both a and n **Therefore, option D is correct.** **Note:** Many students make the mistake of opening the summation value completely by writing the sum of each term one by one which will make calculations more complicated. Keep in mind we use the direct formula of sum of terms to avoid lengthy calculations. Also, if two or more rows of a determinant are the same or zero then its value is zero. Students may mistake while calculating the value of $$\sum\limits_{x = 1}^{n - 1} {3r + 2} $$ The constant term in the summation comes out of it while calculating its value and $$\sum\limits_{r = 1}^{n - 1} 1 = n - 1$$