Question

show that C0+c1+C2+. +c8=256​

Answer 1

Proving of $C_0+C_1+C_2+. +C_8 = 256$

Step-by-step explanation:

We know that,  

$(x+1)^{n} = C_{0}^{n}\, x^{0}+C_{1}^{n}\, x^{1}+C_{2}^{n}\, x^{2}..............C_{n}^{n}\, x^{n}$

Given that n = 8

$(x+1)^{8} = 256\\\\(x+1)^{8} = 2^{8}$

(x+1) = 2

x=2-1

x= 1

Put x = 1 it in equation (x+1)^{n}, we get  

$(1+1)^{8} = C_{0}^{8}\, 1^{0}+C_{1}^{8}\, 1^{1}+C_{2}^{8}\, 1^{2}..............C_{8}^{8}\, 1^{8}$

$(1+1)^{8} = C_{0} +C_{1} +C_{2}+C_{3}+.....+C_{8} = 256$

(2)^8 = 256  

Hence Proved

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