Prove $C_{1} + C_{2} + C_{3} + . . . . . + C_{n} = 2^{n} - 1$

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Solution

Using binomial theorem,
$2^{n} = (1 + 1)^{n}$
$= m \sum k = 0^{n} C_{k} (1)^{n - k} (1)^{k}$
$= n \sum k = 0^{n} C_{k}$
$=^{n} C_{0} +^{n} C_{1} +^{n} C_{2} + . . . \dots \dots \dots . +^{n} C_{n}$
$= 1 +^{n} C_{1} +^{n} C_{2} + . . . \dots \dots . . +^{n} C_{n}$
$∴ 2^{n} - 1 =^{n} C_{1} +^{n} C_{2} + . . . \dots . +^{n} C_{n}$ .

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