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Question

If Dr=∣ ∣ ∣2r12(3r1)4(5r1)xy3z2n13n15n1∣ ∣ ∣
then prove that nr=1Dr=0.

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Solution

nr=12r1=1+2+22+23+.............2n1
= 1.(2n1)21=2n1,
2(3r1)=2(1+3+32+..........+3n1)
=2.1.(3n1)31=3n1,
4(5r1)=4(1+5+52+......+5n1)
=4.1(5n1)51=5n1.
We have used the formula for G.P.
Sn=a(rn1)r1 (r >1)
Rest is as in Q.16

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