If \(\rm D_n=\begin{vmatrix}n&20&30\\\ n^2&40&50\\\ n^3&60&70\end{vmatrix}\) then what is the value of \(\rm \Sigma_{n=1}^4D_n?\)

Concept:

Determinant of a 3×3 Matrix:

• The determinant of a 3×3 matrix can be found by the cofactor expansion method.
• It is expressed as: \( \text{det}(A) = a(ei − fh) − b(di − fg) + c(dh − eg) \)
• If the matrix has a common factor across rows or columns, it can be factored out to simplify calculations.
• \( \sum_{n=1}^{k} f(n) \) is a summation notation indicating summing function values from n = 1 to n = k.
• Important identities used:
  • \( \sum_{n=1}^{k} n = \frac{k(k+1)}{2} \)
  • \( \sum_{n=1}^{k} n^2 = \frac{k(k+1)(2k+1)}{6} \)
  • \( \sum_{n=1}^{k} n^3 = \left( \frac{k(k+1)}{2} \right)^2 \)

Calculation:

Given,

\( D_n = \begin{bmatrix} n & 20 & 30 \\ n^2 & 40 & 50 \\ n^3 & 60 & 70 \\ \end{bmatrix} \)

⇒ Extract common factors 20 and 10:

\( D_n = (20)(10) \begin{bmatrix} n & 1 & 3 \\ n^2 & 2 & 5 \\ n^3 & 3 & 7 \\ \end{bmatrix} \)

⇒ Expand determinant using first column:

\( D_n = 200(-n + 2n^2 - n^3) \)

Now,

\( \sum_{n=1}^{4} D_n = 200 \sum_{n=1}^{4} (-n + 2n^2 - n^3) \)

⇒ \( = 200 \left[ -\frac{4(5)}{2} + 2 \cdot \frac{4(5)(9)}{6} - \left( \frac{4(5)}{2} \right)^2 \right] \)

⇒ \( = 200 (-10 + 60 - 100) \)

∴ The value is -10000.