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Question: If $D_r = \begin{vmatrix} r & 2012 & 2013 \\ 11r^2 & 3r & r \\ r & \frac{1}{r} & \frac{1}{r^2} \end{...

If Dr=∣r2012201311r23rrr1r1r2∣D_r = \begin{vmatrix} r & 2012 & 2013 \\ 11r^2 & 3r & r \\ r & \frac{1}{r} & \frac{1}{r^2} \end{vmatrix}, then ∑r=15Dr\sum_{r=1}^{5}D_r is equal to-

A

0

B

155

C

2013

D

-2012

Answer

0

Explanation

Solution

To solve the problem, we first need to evaluate the determinant DrD_r. The given determinant is: Dr=∣r2012201311r23rrr1r1r2∣D_r = \begin{vmatrix} r & 2012 & 2013 \\ 11r^2 & 3r & r \\ r & \frac{1}{r} & \frac{1}{r^2} \end{vmatrix}

We can expand the determinant along the first row: Dr=r((3r)(1r2)−(r)(1r))−2012((11r2)(1r2)−(r)(r))+2013((11r2)(1r)−(3r)(r))D_r = r \left( (3r) \left(\frac{1}{r^2}\right) - (r) \left(\frac{1}{r}\right) \right) - 2012 \left( (11r^2) \left(\frac{1}{r^2}\right) - (r)(r) \right) + 2013 \left( (11r^2) \left(\frac{1}{r}\right) - (3r)(r) \right)

Simplify each term:

  1. First term: r(3rr2−rr)=r(3r−1)=3−rr \left( \frac{3r}{r^2} - \frac{r}{r} \right) = r \left( \frac{3}{r} - 1 \right) = 3 - r

  2. Second term: −2012(11r2r2−r2)=−2012(11−r2)=−2012×11+2012r2=−22132+2012r2-2012 \left( \frac{11r^2}{r^2} - r^2 \right) = -2012 (11 - r^2) = -2012 \times 11 + 2012r^2 = -22132 + 2012r^2

  3. Third term: 2013(11r2r−3r2)=2013(11r−3r2)=2013×11r−2013×3r2=22143r−6039r22013 \left( \frac{11r^2}{r} - 3r^2 \right) = 2013 (11r - 3r^2) = 2013 \times 11r - 2013 \times 3r^2 = 22143r - 6039r^2

Now, combine these terms to get DrD_r: Dr=(3−r)+(−22132+2012r2)+(22143r−6039r2)D_r = (3 - r) + (-22132 + 2012r^2) + (22143r - 6039r^2)

Group terms by powers of rr: Dr=(2012r2−6039r2)+(−r+22143r)+(3−22132)D_r = (2012r^2 - 6039r^2) + (-r + 22143r) + (3 - 22132) Dr=(2012−6039)r2+(−1+22143)r+(3−22132)D_r = (2012 - 6039)r^2 + (-1 + 22143)r + (3 - 22132) Dr=−4027r2+22142r−22129D_r = -4027r^2 + 22142r - 22129

Next, we need to calculate the sum ∑r=15Dr\sum_{r=1}^{5} D_r: ∑r=15Dr=∑r=15(−4027r2+22142r−22129)\sum_{r=1}^{5} D_r = \sum_{r=1}^{5} (-4027r^2 + 22142r - 22129) This can be written as: ∑r=15Dr=−4027∑r=15r2+22142∑r=15r−22129∑r=151\sum_{r=1}^{5} D_r = -4027 \sum_{r=1}^{5} r^2 + 22142 \sum_{r=1}^{5} r - 22129 \sum_{r=1}^{5} 1

We use the sum formulas for powers of natural numbers: ∑r=1nr=n(n+1)2\sum_{r=1}^{n} r = \frac{n(n+1)}{2} ∑r=1nr2=n(n+1)(2n+1)6\sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6}

For n=5n=5: ∑r=15r=5(5+1)2=5×62=15\sum_{r=1}^{5} r = \frac{5(5+1)}{2} = \frac{5 \times 6}{2} = 15 ∑r=15r2=5(5+1)(2×5+1)6=5×6×116=55\sum_{r=1}^{5} r^2 = \frac{5(5+1)(2 \times 5+1)}{6} = \frac{5 \times 6 \times 11}{6} = 55 ∑r=151=5\sum_{r=1}^{5} 1 = 5

Substitute these values into the sum expression: ∑r=15Dr=−4027(55)+22142(15)−22129(5)\sum_{r=1}^{5} D_r = -4027(55) + 22142(15) - 22129(5)

Calculate the products: −4027×55=−221485-4027 \times 55 = -221485 22142×15=33213022142 \times 15 = 332130 −22129×5=−110645-22129 \times 5 = -110645

Now, sum these values: ∑r=15Dr=−221485+332130−110645\sum_{r=1}^{5} D_r = -221485 + 332130 - 110645 ∑r=15Dr=332130−(221485+110645)\sum_{r=1}^{5} D_r = 332130 - (221485 + 110645) ∑r=15Dr=332130−332130\sum_{r=1}^{5} D_r = 332130 - 332130 ∑r=15Dr=0\sum_{r=1}^{5} D_r = 0

The final answer is 0\boxed{0}.