Question

Let $D_{k}=\begin{vmatrix}a & 2^{k} & 2^{16}-1 \\ b & 3\left(4^{k}\right) & 2\left(4^{16}-1\right) \\ c & 7\left(8^{k}\right) & 4\left(8^{16}-1\right)\end{vmatrix}$, then the value of $\displaystyle\sum_{k=1}^{16} D_{k}$ is

• A. 0
• B. a + b + c
• C. ab + bc + ca
• D. None of these

Answer 1

Answer: (A)

We have, $ \sum\limits_{k=1}^{16}{{{D}_{k}}}=\left| \begin{matrix} a & \left( \sum\limits_{k=1}^{16}{{{2}^{k}}} \right) & {{2}^{16}}-1 \\ b & 3\left( \sum\limits_{k=1}^{16}{{{4}^{4}}} \right) & 2({{4}^{16}}-1) \\ c & 7\left( \sum\limits_{k=1}^{16}{{{8}^{k}}} \right) & 4({{8}^{16}}-1) \\ \end{matrix} \right| $
$ \Rightarrow $ $ \sum\limits_{k=1}^{16}{{{D}_{k}}}=\left| \begin{matrix} a & \left( \frac{{{2}^{16}}-1}{2-1} \right) & {{2}^{16}}-1 \\ b & 3\times 4\left( \frac{{{4}^{16}}-1}{4-1} \right) & 2({{4}^{16}}-1) \\ c & 7\times 8\left( \frac{{{8}^{16-1}}}{8-1} \right) & 4({{8}^{16}}-1) \\ \end{matrix} \right| $
$ \Rightarrow $ $ \sum\limits_{k=1}^{16}{{{D}_{k}}}=\left| \begin{matrix} a & 2({{2}^{16}}-1) & {{2}^{16}}-1 \\ b & 4({{4}^{16}}-1) & 2({{4}^{16}}-1) \\ c & 8({{8}^{16}}-1) & 4({{8}^{16}}-1) \\ \end{matrix} \right| $
$ \Rightarrow $ $ \sum\limits_{k=1}^{16}{{{D}_{k}}}=2\left| \begin{matrix} a & ({{2}^{16}}-1) & {{2}^{16}}-1 \\ b & 2({{4}^{16}}-1) & 2({{4}^{16}}-1) \\ c & 4({{8}^{16}}-1) & 4({{8}^{16}}-1) \\ \end{matrix} \right| $
[taking $2$ common from $ {{C}_{2}} $ ]
$ \Rightarrow $ $ \sum\limits_{k=1}^{16}{{{D}_{k}}=2\times 0=0} $