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![[Post New]](/templates/default/images/icon_minipost_new.gif) 28 Mar 2008 19:28:22 IST
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20C0 + 20C1 + ......+ 20C9 + 20C10 = ?
Please show the procedure as well.
Has the sum of coefficients being 2n has something to do with it?
perhaps 219 + 20C0?
Is this right??
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28 Mar 2008 19:31:54 IST
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note that 20C9 = 20C11 and so on so the sum is 1/2(20C0 + 20C1........20C20) + 20C10/2 implies its is 2^20/2 + (20C10)/2 = 2^19 + 20C10/2
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(1+x)^20 = 20C0 + 20C1*x +............. 20C20* x^2
x=1
2^20 = 20c0 + 20c1 +.........+20c11............. + 20c20
2^20= 20c0 + 20c1+............ 20c10 + (20c9 + 20c8 +.......20c0)
2^20 = 2*(20c0+20c1+..... 20c10) - 20c10
(2*20 + 20c10)/2 = 20c0+ 20c1+.... +20c10
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28 Mar 2008 19:44:27 IST
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Thank you!
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