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Set Sn=1+q+q2++qn and Tn=1+q+12+q+122++q+12nwhere q is a real number and q1 If 101C1+101C2S1++101 C101S100=αT100 , then α is equal to 

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a
2100
b
299
c
202
d
200
 

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detailed solution

Correct option is A

Here,  101C1+101C2S1++101C101S100=αT100

101C1+101C2(1+q)+101C31+q+q2++101C1011+q++q100

=2α11+q2101(1q)101C1(1q)+101C21q2++101C1011q101=2α11+q210121011(1+q)1011=2α11+q2101210111+q2101=2α11+q2101α=2100

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