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Questions  

Let A=1    a0    1, where a>0, Sum of  the

series S = trace (A) + trace12A+trace122A2+trace123A3+ … is

a
3
b
4
c
6
d
8

detailed solution

Correct option is B

A=I+B, where B=0    a0    0B2=O we get Br=O ∀r≥2Ar=(I+B)r=I+rB=1ra01 ∀r≥1trace 12rAr=12r(2)=12r−1 ∀r≥1.S=2+1+122+123+⋯=21−12=4

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