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The sum of the series 20C020C1 + 20C220C3 + ...... + 20C10 is

(a) 20C10 

(b)  \(\frac{1}{2}\)20C10 

(c) 0 

(d) 20C15

1 Answer

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Answer : (b)  \(\frac{1}{2}\)20C10

(1 + x) 20 = 20C0 + 20C1 x + 20C2 x2 + 20C3 x3 + ..... + 20C19 x19 + 20C20 x20

On putting x = – 1, we get

0 = 20C020C1 + 20C2 20C3 + ..... – 20C19 + 20C20

⇒ 0 = 20C0 20C1 + ..... – 20C9 + 20C1020C9 + 20C8 .... + 20C

(∵ nCr = nCn–r )

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