prove that c0+c1+c2+ +cn=2n

from the binomial expansion : (1+x)n=C0n+C1n x+C2n x2+ +Cnn xn (1)now substituting x =1 on the both sides , we have:(1+1)n=C0n+C1n 1+C2n 12+ +Cnn 1n⇒C0n+C1n+C2n+ +Cnn=2n hope this helps you

prove that c₀+c₁+c₂+........+c_n=2ⁿ

from the binomial expansion :

(1 + x)^{n} = {}^{n}C_{0} + {}^{n}C_{1} . x + {}^{n}C_{2} . x^{2} + . . . . . . . . . . . + {}^{n}C_{n} . x^{n} . . . . . . . . . . . (1) n o w s u b s t i t u t i n g x = 1 o n t h e b o t h s i d e s, w e h a v e : (1 + 1)^{n} = {}^{n}C_{0} + {}^{n}C_{1} . 1 + {}^{n}C_{2} . 1^{2} + . . . . . . . . + {}^{n}C_{n} . 1^{n} \Rightarrow {}^{n}C_{0} + {}^{n}C_{1} + {}^{n}C_{2} + . . . . + {}^{n}C_{n} = 2^{n}

hope this helps you

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prove that c0+c1+c2+........+cn=2n

prove that c₀+c₁+c₂+........+c_n=2ⁿ