Question

How is nC0+nC1+nC2+....+nCn=2n

Answer 1

For every term we have two choices- yes/no therefore for n terms we have 2x2x2x2.....n = 2^n

Answer 2

This is true for any positive integers a, b and n ↓↓↓

$(a+b)^n =\ ^n\!C_0 \ a^n +\ ^n \! C_1 \ a^{n-1} b+ \ ^n\! C_2\ a^{n-2}b^2+...+\ ^n\! C_{n-1} \ ab^{n-1}+\ ^n\! C_n \ b^n$

Actually the question is to prove,

$^n\! C_0+\ ^n\!C_1+\ ^n\!C_2+...+\ ^n\! C_n=2^n$

Mistake is that the RHS is 2ⁿ, not 2n.

Proof:

$\Large \textit{RHS...} \\ \\ \\ \Rightarrow\ 2^n \\ \\ \Rightarrow\ (1+1)^n \\ \\ \Rightarrow\ ^n\!C_0\cdot 1^n+\ ^n\!C_1\cdot 1^{n-1} \cdot 1+\ ^n\! C_2\cdot 1^{n-2}\cdot 1^2+...+\ ^n\! C_n\cdot 1^n \\ \\ \Rightarrow\ ^n\! C_0+\ ^n\!C_1+\ ^n\!C_2+...+\ ^n\! C_n \\ \\ \\ \Rightarrow\ \Large \textit{...LHS}$

Hence Proved!