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Question

If Dk=∣ ∣ ∣ ∣ ∣ ∣2k11k(k+1)sinkθxyz2n1nn+1sin(n+12)θsinn2θsinθ/2∣ ∣ ∣ ∣ ∣ ∣ then nk=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 z
The (1,1) term,=Σ2k1
=20+21+22+...2n1 (Geometric series)
=(2n1)
The (1,2) term,=Σ1k(k+1)=Σ(1k1k+1)
=11n+1
=nn+1
The (1,3) term,
=Σsinkθ=sin(n+12)θsinn2θsinθ2
As R1=R3 so, Δ=0 and is independent of any parameters.

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