If ${D_P} = \left| {\begin{array}{*{20}{c}}
  P&{15}&8 \\
  {{P^2}}&{35}&9 \\
  {{P^3}}&{25}&{10}
\end{array}} \right|$, then ${D_1} + {D_2} + {D_3} + {D_4} + {D_5}$ is equal to
A. -29000
B. -25000
C. 25000
D. None of these

Hint:To find the value of ${D_1} + {D_2} + {D_3} + {D_4} + {D_5}$, we first need to solve determinant ${D_P} = \left| {\begin{array}{*{20}{c}}
  P&{15}&8 \\
  {{P^2}}&{35}&9 \\
  {{P^3}}&{25}&{10}
\end{array}} \right|$ and find its value. After finding its value in terms of P, we need to substitute $P = 1,2,3,4,5$ and add all those values to get our answer.

Complete step by step answer:
In this question, we are given a determinant with variable P and we need to find out the value of ${D_1} + {D_2} + {D_3} + {D_4} + {D_5}$.
Given determinant: ${D_P} = \left| {\begin{array}{*{20}{c}}
  P&{15}&8 \\
  {{P^2}}&{35}&9 \\
  {{P^3}}&{25}&{10}
\end{array}} \right|$
Now, first of all we will find the value of the given determinant and then substitute the value of P with 1, 2, 3, 4 and 5 one by one and add those values.
Therefore, we get
${D_P} = \left| {\begin{array}{*{20}{c}}
  P&{15}&8 \\
  {{P^2}}&{35}&9 \\
  {{P^3}}&{25}&{10}
\end{array}} \right| = P\left| {\begin{array}{*{20}{c}}
  {35}&9 \\
  {25}&{10}
\end{array}} \right| - 15\left| {\begin{array}{*{20}{c}}
  {{P^2}}&9 \\
  {{P^3}}&{10}
\end{array}} \right| + 8\left| {\begin{array}{*{20}{c}}
  {{P^2}}&{35} \\
  {{P^3}}&{25}
\end{array}} \right|$
${D_P} = P\left( {35 \times 10 - 25 \times 9} \right) - 15\left( {{P^2} \times 10 - {P^3} \times 9} \right) + 8\left( {{P^2} \times 25 - {P^3} \times 35} \right) \\
\Rightarrow {D_P} = P\left( {350 - 225} \right) - 15\left( {10{P^2} - 9{P^3}} \right) + 8\left( {25{P^2} - 35{P^3}} \right) \\
\Rightarrow {D_P} = 125P - 150{P^2} + 135{P^3} + 200{P^2} - 280{P^3} \\
\Rightarrow {D_P} = 125P + 50{P^2} - 145{P^3} \\ $
Hence, we have found the value of the given determinant and now we need to substitute $P = 1,2,3,4,5$ and add those results.

Therefore, For $P = 1$:
${D_P} = 125P + 50{P^2} - 145{P^3} \\
\Rightarrow {D_1} = 125\left( 1 \right) + 50{\left( 1 \right)^2} - 145{\left( 1 \right)^3} \\
\Rightarrow {D_1} = 125 + 50 - 145 \\
\Rightarrow {D_1} = 30 \\ $
For $P = 2$:
${D_P} = 125P + 50{P^2} - 145{P^3} \\
\Rightarrow {D_2} = 125\left( 2 \right) + 50{\left( 2 \right)^2} - 145{\left( 2 \right)^3} \\
\Rightarrow {D_2} = 250 + 200 - 1160 \\
\Rightarrow {D_2} = - 710 \\ $
For $P = 3$:
${D_P} = 125P + 50{P^2} - 145{P^3} \\
\Rightarrow {D_3} = 125\left( 3 \right) + 50{\left( 3 \right)^2} - 145{\left( 3 \right)^3} \\
\Rightarrow {D_3} = 375 + 450 - 3915 \\
\Rightarrow {D_3} = - 3090 \\ $
For $P = 4$:
${D_P} = 125P + 50{P^2} - 145{P^3} \\
\Rightarrow {D_4} = 125\left( 4 \right) + 50{\left( 4 \right)^2} - 145{\left( 4 \right)^3} \\
\Rightarrow {D_4} = 500 + 800 - 9280 \\
\Rightarrow {D_4} = - 7980 \\ $
For $P = 5$:
${D_P} = 125P + 50{P^2} - 145{P^3} \\
\Rightarrow {D_5} = 125\left( 5 \right) + 50{\left( 5 \right)^2} - 145{\left( 5 \right)^3} \\
\Rightarrow {D_5} = 625 + 1250 - 18125 \\
\Rightarrow {D_5} = - 16250 \\ $
Therefore, we have all the values we need and now we need to just add them. Therefore, we get
${D_1} + {D_2} + {D_3} + {D_4} + {D_5} = 30 - 710 - 3090 - 7980 - 16250 \\
\therefore {D_1} + {D_2} + {D_3} + {D_4} + {D_5} = - 28000 \\ $
Hence, the correct answer is option D.

Note: We can also solve this question using the following method.
$\Rightarrow {D_P} = \left| {\begin{array}{*{20}{c}}
  P&{15}&8 \\
  {{P^2}}&{35}&9 \\
  {{P^3}}&{25}&{10}
\end{array}} \right|$
Substitute $P = 1,2,3,4,5$. Therefore, we get
For $P = 1$:
$\Rightarrow {D_1} = \left| {\begin{array}{*{20}{c}}
  1&{15}&8 \\
  {{1^2}}&{35}&9 \\
  {{1^3}}&{25}&{10}
\end{array}} \right| \\
\Rightarrow {D_1} = 1\left( {350 - 225} \right) - 15\left( {10 - 9} \right) + 8\left( {25 - 35} \right) \\
\Rightarrow {D_1} = 125 - 15 - 80 \\
\Rightarrow {D_1} = 30 \\ $
For $P = 2$:
$\Rightarrow {D_2} = \left| {\begin{array}{*{20}{c}}
  2&{15}&8 \\
  {{2^2}}&{35}&9 \\
  {{2^3}}&{25}&{10}
\end{array}} \right| \\
\Rightarrow {D_2} = 2\left( {350 - 225} \right) - 15\left( {40 - 72} \right) + 8\left( {100 - 280} \right) \\
\Rightarrow {D_2} = 250 + 480 - 1440 \\
\Rightarrow {D_2} = - 710 \\ $
For $P = 3$:
$\Rightarrow {D_3} = \left| {\begin{array}{*{20}{c}}
  3&{15}&8 \\
  {{3^2}}&{35}&9 \\
  {{3^3}}&{25}&{10}
\end{array}} \right| \\
\Rightarrow {D_3} = 3\left( {350 - 225} \right) - 15\left( {90 - 243} \right) + 8\left( {225 - 945} \right) \\
\Rightarrow {D_3} = 375 + 2295 - 5760 \\
\Rightarrow {D_3} = - 3090 \\ $
For $P = 4$:
$\Rightarrow {D_4} = \left| {\begin{array}{*{20}{c}}
  4&{15}&8 \\
  {{4^2}}&{35}&9 \\
  {{4^3}}&{25}&{10}
\end{array}} \right| \\
\Rightarrow {D_4} = 4\left( {350 - 225} \right) - 15\left( {160 - 576} \right) + 8\left( {400 - 2240} \right) \\
\Rightarrow {D_4} = 500 + 6240 - 14720 \\
\Rightarrow {D_4} = - 7980 \\ $
For $P = 5$:
$\Rightarrow {D_5} = \left| {\begin{array}{*{20}{c}}
  5&{15}&8 \\
  {{5^2}}&{35}&9 \\
  {{5^3}}&{25}&{10}
\end{array}} \right| \\
\Rightarrow {D_5} = 5\left( {350 - 225} \right) - 15\left( {250 - 1125} \right) + 8\left( {625 - 4375} \right) \\
\Rightarrow {D_5} = 625 + 13125 - 30000 \\
\Rightarrow {D_5} = - 16250 \\ $
Therefore,
$\Rightarrow {D_1} + {D_2} + {D_3} + {D_4} + {D_5} = 30 - 710 - 3090 - 7980 - 16250 \\
\therefore {D_1} + {D_2} + {D_3} + {D_4} + {D_5} = - 28000 \\ $