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Question

IF Sn=nr=01nCr and tn=nr=0rnCr, then tnSn is equal to

A
2n12
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B
12n1
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C
n1
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D
12n
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Solution

The correct option is D 12n
Sn=1nC0+1nC1+1nC2++1nCntn=0nC0+1nC1+2nC2++nnCntn=nnCn+n1nCn1+n2nCn2++0nC0
Add, 2tn=(n)[1nC0+1nC1+1nCn]=nSntnSn=n2

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