Question

Let $D_k=\left|\begin{array}{ccc}a& 2^k& 2^{16}-1\\ b& 3\left(4^k\right)& 2(4^{16}-1)\\ c& 7\left(8^k\right)& 4(8^{16}-1)\end{array}\right|$, then the value of $\sum _{k=1}^{16}{D}_k$ is

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

Answer 1

The correct option is A $0$

$D_k=\left|\begin{array}{ccc}a& 2^k& 2^{16}-1\\ b& 3\left(4^k\right)& 2(4^{16}-1)\\ c& 7\left(8^k\right)& 4(8^{16}-1)\end{array}\right|$

$\sum _{k=1}^{16}{D}_k=\left|\begin{array}{ccc}a& {\sum }_{k=1}^{16}{2}^k& 2^{16}-1\\ b& 3\left({\sum }_{k=1}^{16}{4}^k\right)& 2(4^{16}-1)\\ c& 7\left({\sum }_{k=1}^{16}{8}^k\right)& 4(8^{16}-1)\end{array}\right|$

$\sum _{k=1}^{16}{D}_k=\left|\begin{array}{ccc}a& 2\left(\frac{2^{16}-1}{2-1}\right)& 2^{16}-1\\ b& 3\left\{4\left(\frac{4^{16}-1}{4-1}\right)\right\}& 2(4^{16}-1)\\ c& 7\left\{8\left(\frac{8^{16}-1}{8-1}\right)\right\}& 4(8^{16}-1)\end{array}\right|$

$\sum _{k=1}^{16}{D}_k=\left|\begin{array}{ccc}a& 2(2^{16}-1)& 2^{16}-1\\ b& 4(4^{16}-1)& 2(4^{16}-1)\\ c& 8(8^{16}-1)& 4(8^{16}-1)\end{array}\right|$

$\sum _{k=1}^{16}{D}_k=2\left|\begin{array}{ccc}a& (2^{16}-1)& 2^{16}-1\\ b& 2(4^{16}-1)& 2(4^{16}-1)\\ c& 4(8^{16}-1)& 4(8^{16}-1)\end{array}\right|$

Two columns of the determinant are same

Therefore, $\sum _{k=1}^{16}{D}_k=0$

So, option A is correct.