A) 0
B) a + b + c
C) ab + 6c + ca
D) None of these
Correct Answer: A
Solution :
We have, \[\sum\limits_{k=1}^{16}{{{D}_{k}}}=\left| \begin{matrix} a & \left( \sum\limits_{k=1}^{16}{{{2}^{k}}} \right) & {{2}^{16}}-1 \\ b & 3\left( \sum\limits_{k=1}^{16}{{{4}^{4}}} \right) & 2({{4}^{16}}-1) \\ c & 7\left( \sum\limits_{k=1}^{16}{{{8}^{k}}} \right) & 4({{8}^{16}}-1) \\ \end{matrix} \right|\] \[\Rightarrow \]\[\sum\limits_{k=1}^{16}{{{D}_{k}}}=\left| \begin{matrix} a & \left( \frac{{{2}^{16}}-1}{2-1} \right) & {{2}^{16}}-1 \\ b & 3\times 4\left( \frac{{{4}^{16}}-1}{4-1} \right) & 2({{4}^{16}}-1) \\ c & 7\times 8\left( \frac{{{8}^{16-1}}}{8-1} \right) & 4({{8}^{16}}-1) \\ \end{matrix} \right|\] \[\Rightarrow \]\[\sum\limits_{k=1}^{16}{{{D}_{k}}}=\left| \begin{matrix} a & 2({{2}^{16}}-1) & {{2}^{16}}-1 \\ b & 4({{4}^{16}}-1) & 2({{4}^{16}}-1) \\ c & 8({{8}^{16}}-1) & 4({{8}^{16}}-1) \\ \end{matrix} \right|\] \[\Rightarrow \]\[\sum\limits_{k=1}^{16}{{{D}_{k}}}=2\left| \begin{matrix} a & ({{2}^{16}}-1) & {{2}^{16}}-1 \\ b & 2({{4}^{16}}-1) & 2({{4}^{16}}-1) \\ c & 4({{8}^{16}}-1) & 4({{8}^{16}}-1) \\ \end{matrix} \right|\] [talong 2 common from \[{{C}_{2}}\]] \[\Rightarrow \]\[\sum\limits_{k=1}^{16}{{{D}_{k}}=2\times 0=0}\]
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