Alg10.10

From Exampleproblems

Prove that ${C \choose 1}+{C \choose 2}+{C \choose 4}+...=2^{n-1}={C \choose 1}+{C \choose 3}+{C \choose 5}+.......\,$

$(1+a)^n={C \choose 0}+{C \choose 1}a+{C \choose 2}a^2+.....\,$

Put $a=-1\,$ in the above,

$(1-1)^n={C \choose 0}-{C \choose 1}+{C \choose 2}-{C \choose 3}+.......\,$

${C \choose 0}+{C \choose 2}+{C \choose 4}+......={C \choose 1}+{C \choose 3}+{C \choose 5}+.......\,$

Recall that $a=b=\frac{a+b}{2}\,$

Therefore ${C \choose 0}+{C \choose 2}+{C \choose 4}+......={C \choose 1}+{C \choose 3}+{C \choose 5}+.......=\frac{{C \choose 0}+{C \choose 1}+{C \choose 2}+{C \choose 3}+........}{2}=\frac{2^n}{2}\,$

Hence ${C \choose 1}+{C \choose 2}+{C \choose 4}+...=2^{n-1}={C \choose 1}+{C \choose 3}+{C \choose 5}+.......\,$

Main Page:Algebra

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