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Question

Prove that
10k=020Ck=219+12 20C10

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Solution

Let S=10k=020Ck=20C0+20C1+20C2+....+20C10....(1)
In the binomial expansion
(1+x)20=20C0+20C1x+20C2x2+20C3x3+.....20Cx20
at x=1
220=20C0+20C1+20C2+.....+20C20
As nCr=nCnr (property)
220C0=20C20,20C1=20C19,20C2=20C18,....etc
220+20C10=220C0+220C1+220C2+.....220C9+220C10
220+20C10=2[20C0+20C1+20C2+....+20C10]
220+20C10=2SS=12[220+20C10]=219+1220C10

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