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# Let ‘m’ be a positive integer, a, b, c are three real numbers and${{\Delta }_{r}}=\left| \begin{matrix} 2r-1 & {}^{m}{{C}_{r}} & 1 \\ {{m}^{2}}-1 & {{2}^{m}} & m+1 \\ a & b & c \\\end{matrix} \right|$$\left( 0\le r\le m \right)$ Then the value of $\sum\limits_{r=0}^{m}{{{\Delta }_{r}}}$ is given by(a) 0(b) ${{m}^{2}}-1$(c) ${{2}^{m}}abc$(d) None of the above.

Last updated date: 20th Jun 2024
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Hint: We solve this problem by using the simple formula for summation of determinant that is we apply the summation of determinant to inside such that it applies to all rows that are having the variable of summation that is if
$\Delta =\left| \begin{matrix} f\left( x \right) & g\left( x \right) & h\left( x \right) \\ p & q & r \\ a & b & c \\ \end{matrix} \right|$
Then the summation with respect to $'x'$ is given as
$\Rightarrow \sum\limits_{x=0}^{n}{\Delta }=\left| \begin{matrix} \sum\limits_{x=0}^{n}{f\left( x \right)} & \sum\limits_{x=0}^{n}{g\left( x \right)} & \sum\limits_{x=0}^{n}{h\left( x \right)} \\ p & q & r \\ a & b & c \\ \end{matrix} \right|$
By using the above formula we get the required determinant by using the summation of first $'n'$ whole numbers as
$\sum\limits_{r=0}^{n}{r}=\dfrac{n\left( n+1 \right)}{2}$
$\sum\limits_{r=0}^{n}{1}=n+1$

We are given that
${{\Delta }_{r}}=\left| \begin{matrix} 2r-1 & {}^{m}{{C}_{r}} & 1 \\ {{m}^{2}}-1 & {{2}^{m}} & m+1 \\ a & b & c \\ \end{matrix} \right|$
We are asked to find the summation of above determinant with respect to $'r'$
We know that if
$\Delta =\left| \begin{matrix} f\left( x \right) & g\left( x \right) & h\left( x \right) \\ p & q & r \\ a & b & c \\ \end{matrix} \right|$
Then the summation with respect to $'x'$ is given as
$\Rightarrow \sum\limits_{x=0}^{n}{\Delta }=\left| \begin{matrix} \sum\limits_{x=0}^{n}{f\left( x \right)} & \sum\limits_{x=0}^{n}{g\left( x \right)} & \sum\limits_{x=0}^{n}{h\left( x \right)} \\ p & q & r \\ a & b & c \\ \end{matrix} \right|$
By using the above formula to given determinant we get
$\Rightarrow \sum\limits_{r=0}^{m}{{{\Delta }_{r}}}=\left| \begin{matrix} \sum\limits_{r=0}^{m}{\left( 2r-1 \right)} & \sum\limits_{r=0}^{m}{{}^{m}{{C}_{r}}} & \sum\limits_{r=0}^{m}{1} \\ {{m}^{2}}-1 & {{2}^{m}} & m+1 \\ a & b & c \\ \end{matrix} \right|..........equation(i)$
Now, let us assume that the first row first element as
$\Rightarrow {{a}_{1}}=\sum\limits_{r=0}^{m}{\left( 2r-1 \right)}$
We know that the summation of first $'n'$ whole numbers as
$\sum\limits_{r=0}^{n}{r}=\dfrac{n\left( n+1 \right)}{2}$
$\sum\limits_{r=0}^{n}{1}=n+1$

By using these formulas to above equation we get
\begin{align} & \Rightarrow {{a}_{1}}=2\left( \dfrac{m\left( m+1 \right)}{2} \right)-\left( m+1 \right) \\ & \Rightarrow {{a}_{1}}={{m}^{2}}+m-m-1 \\ & \Rightarrow {{a}_{1}}={{m}^{2}}-1..........equation(ii) \\ \end{align}
Now let us assume that second element of first row as
\begin{align} & \Rightarrow {{a}_{2}}=\sum\limits_{r=0}^{m}{{}^{m}{{C}_{r}}} \\ & \Rightarrow {{a}_{2}}={}^{m}{{C}_{0}}+{}^{m}{{C}_{1}}+{}^{m}{{C}_{2}}+...........+{}^{m}{{C}_{m}} \\ \end{align}
We know that binomial expansion as’
${{\left( 1+x \right)}^{n}}={}^{n}{{C}_{0}}+{}^{n}{{C}_{1}}x+{}^{n}{{C}_{2}}{{x}^{2}}+.........+{}^{n}{{C}_{n}}{{x}^{n}}$
Now by substituting $x=1$ in above equation we get
$\Rightarrow {{2}^{n}}={}^{n}{{C}_{0}}+{}^{n}{{C}_{1}}+{}^{n}{{C}_{2}}+.........+{}^{n}{{C}_{n}}$
By using this result in the second element of first row we get
$\Rightarrow {{a}_{2}}={{2}^{m}}.......equation(iii)$
Let us assume that the third element of first row we get
$\Rightarrow {{a}_{3}}=\sum\limits_{r=0}^{m}{1}$
We know that the standard result to summation as
$\sum\limits_{r=0}^{n}{1}=n+1$
By using this result to above equation we get
$\Rightarrow {{a}_{3}}=m+1.........equation(iv)$
Now, by substituting the equation (ii), equation (iii) and equation (iv) in equation (i) we get
$\Rightarrow \sum\limits_{r=0}^{m}{{{\Delta }_{r}}}=\left| \begin{matrix} {{m}^{2}}-1 & {{2}^{m}} & m+1 \\ {{m}^{2}}-1 & {{2}^{m}} & m+1 \\ a & b & c \\ \end{matrix} \right|$

Here, we can see that there are two equal rows.
We know that the determinant having two equal rows or columns is zero.
By using the result we get
$\Rightarrow \sum\limits_{r=0}^{m}{{{\Delta }_{r}}}=0$
Therefore the summation of given determinant with respect to $'r'$ is given as
$\therefore \sum\limits_{r=0}^{m}{{{\Delta }_{r}}}=0$

So, the correct answer is “Option A”.

Note: Students may do mistake in the summation of numbers.
Here we have the formula of the summation of first $'n'$ whole numbers as
$\sum\limits_{r=0}^{n}{1}=n+1$
Here, the summation stars from 0 so, we get total of $n+1$ terms.
But students may do mistake and take the formula as
$\sum\limits_{r=0}^{n}{1}=n$