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Question

# If Dk=∣∣ ∣ ∣ ∣ ∣ ∣∣2k−11k(k+1)sinkθxyz2n−1nn+1sin(n+12)θsinn2θsinθ/2∣∣ ∣ ∣ ∣ ∣ ∣∣ then n∑k=1Dk is equal to

A
0
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B
independent of n
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C
independent of θ
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D
independent of x, y and z
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Solution

## The correct options are A 0 B independent of n C independent of θ D independent of x, y and zThe (1,1) term,=Σ2k−1=20+21+22+...2n−1 (Geometric series)=(2n−1)The (1,2) term,=Σ1k(k+1)=Σ(1k−1k+1)=1−1n+1=nn+1The (1,3) term,=Σsinkθ=sin(n+12)θsinn2θsinθ2As R1=R3 so, Δ=0 and is independent of any parameters.

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