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

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 

  1. A

    2100

  2. B

    299

  3. C

    202

  4. D

    200

     

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    Solution:

    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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