Q.

Let Sk=3k  100C0  100Ck3k1  100C1  99Ck1+3k2  100C2  98Ck2+...+1k  100Ck  100kC0,
  then select correct alternative(s) k=0,1,2,.....,100

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a

S1+S3+S5+....+S99>S2+S4+....+S100

b

S67 is the greatest amongst these 101 numbers

c

 S66 is the greatest amongst these 101 numbers

d

S1+S3+S5+....+S99S2+S4+....+S100

answer is A, B.

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Detailed Solution

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Sk=r=0k3kr   100Cr   100rCkr1r Sk=r=0k3k   100Ck kCr13r Sk=2k   100Ck

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