Question

If $D_P=\left|\begin{array}{ccc}P& 15& 8\\ P^2& 35& 9\\ P^3& 25& 10\end{array}\right|,$ then $D_1+D_2+D_3+D_4+D_5$ is equal to -

• A. $-29000$
• B. $-25000$
• C. $25000$
• D. none of these

Answer 1

The correct option is A $-29000$

Given

$D_p=\left|\begin{array}{ccc}p& 15& 8\\ p^2& 35& 9\\ p^3& 25& 10\end{array}\right|$

$=p(350-225)-15(10p^2-9p^3)+8(25p^2-35p^3)$

$D_1=1(350-225)-15\left(10\right(1{)}^2-9\left(1{)}^3\right)+8\left(25\right(1{)}^2-35\left(1{)}^3\right)$

$=125-15.\left(1\right)+8(-10)$

$=125-80-15$

$=125-95$

$=30$

$D_2=2(350-225)-15\left(10\right(2{)}^2-9\left(2{)}^3\right)+8\left(25\right(2{)}^2-35\left(2{)}^3\right)$

$=2.\left(125\right)-15(40-72)+8(100-280)$

$=250+15\left(32\right)-1440$

$=730-1440$

$=-710$

$D_3=3(350-225)-15\left(10\right(3{)}^2-9\left(3{)}^3\right)+8\left(25\right(3{)}^2-35\left(3{)}^3\right)$

$=3.\left(125\right)-15(90-243)+8(225-945)$

$=375+15\left(153\right)-8\left(720\right)$

$=375+2,295-5760$

$=-3090$.

$D_4=4(350-225)-15\left(10\right(4{)}^2-9\left(4{)}^3\right)+8\left(25\right(4{)}^2-35\left(4{)}^3\right)$

$=4.\left(125\right)-15(160-576)+8(400-2240)$

$=500+6240-8.\left(1840\right)$

$=5740-14720$

$=-8980$

$D_5=5(350-225)-15\left(10\right(5{)}^2-9\left(5{)}^3\right)+8\left(25\right(5{)}^2-35\left(5{)}^3\right)$

$=5\left(125\right)-15(250-1125)+8(625-4375)$

$=625+13125+(-30,000)$

$=-16250$.

$\therefore D_1+D_2+D_3+D_4+D_5=30-710-3090-8980-16250$

$=-29000$.