If Dk=3k1(k+1)(k+2)cos⁡(k+1)θ3n−12nn+1sin⁡nθ2cos⁡(n−1)θ2sin⁡θ2abc then ∑k=0n−1 Dk is

∑k=0n−1 3k=1+3+32+….+3n−1=3n−13−1=3n−12⇒∑k=0n−1 1(k+1)(k+2)=∑k=0n−1 1k+1−1k+2=nn+1⇒∑k=0n−1 cos⁡(k+1)θ=1+cos⁡θ+….+cos⁡(n−1)θRe⁡1+eiθ+ei2θ+…..+ei(n−1)θ=sin⁡nθ2cos⁡(n−1)θ2sin⁡θ2

If Dk=3k1(k+1)(k+2)cos⁡(k+1)θ3n−12nn+1sin⁡nθ2cos⁡(n−1)θ2sin⁡θ2abc then ∑k=0n−1 Dk is

∑k=0n−1 3k=1+3+32+….+3n−1=3n−13−1=3n−12⇒∑k=0n−1 1(k+1)(k+2)=∑k=0n−1 1k+1−1k+2=nn+1⇒∑k=0n−1 cos⁡(k+1)θ=1+cos⁡θ+….+cos⁡(n−1)θRe⁡1+eiθ+ei2θ+…..+ei(n−1)θ=sin⁡nθ2cos⁡(n−1)θ2sin⁡θ2

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If $D_{k} = |\begin{array}{ccc} 3^{k} & \frac{1}{(k + 1) (k + 2)} & \cos (k + 1) θ \\ \frac{3^{n} - 1}{2} & \frac{n}{n + 1} & \frac{\sin \frac{nθ}{2} \cos \frac{(n - 1) θ}{2}}{\sin \frac{θ}{2}} \\ a & b & c \end{array}|$ then $\sum_{k = 0}^{n - 1} D_{k}$ is

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independent of n

independent of a,b,c

$a + b + c$

$a + 2 b + c$

answer is B.

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Detailed Solution

$\begin{array}{l} \sum_{k = 0}^{n - 1} 3^{k} = 1 + 3 + 3^{2} + \dots . + 3^{n - 1} = \frac{3^{n} - 1}{3 - 1} = \frac{3^{n} - 1}{2} \\ \Rightarrow \sum_{k = 0}^{n - 1} \frac{1}{(k + 1) (k + 2)} = \sum_{k = 0}^{n - 1} (\frac{1}{k + 1} - \frac{1}{k + 2}) = \frac{n}{n + 1} \\ \Rightarrow \sum_{k = 0}^{n - 1} \cos (k + 1) θ = 1 + \cos θ + \dots . + \cos (n - 1) θ \\ Re [1 + e^{iθ} + e^{i 2 θ} + \dots . . + e^{i (n - 1) θ}] = \frac{\sin \frac{nθ}{2} \cos \frac{(n - 1) θ}{2}}{\sin \frac{θ}{2}} \end{array}$