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Question

The sum of the series 10r=020Cr is

A
21912.20C10
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B
219+12.20C10
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C
219
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D
220
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Solution

The correct option is B 219+12.20C10
We have, 10r=020Cr=20C0+20C1+...+20C10
But 20C0+20C1+...+20C20=220
and 20C20=20C0;20C19=20C1;20C18=20C2... and 20C11=20C9
10r=020Cr=(20C0+20C1+...+20C20)(20C11+20C12+...+20C20)=220+20C10(20C10+20C9+...+20C0)2[20C0+20C1+...+20C10]=220+20C10
20C0+20C1+...+20C10=219+12.20C10

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