Question

How can you prove that `1/(0!)+1/(1!)(n-1)+1/(2!)(n-1)(n-2)+1/(3!)(n-1)(n-2)(n-3)+... = 2^(n-1)` ?

Answer 1

Please see below.

Explanation:

Binomial expansion of `(1+x)^m`

= `C_0^m+C_1^mx+C_2^mx^2+C_3^mx^3+..+C_m^nx^m`

where `C_r^m=(m(m-1)(m-2)..(m-r+1))/(1*2*3*......*r)` and `C_0^m=1/(0!)=1`

Hence `(1+1)^((n-1))`

= `C_0^(n-1)+C_1^(n-1)+C_2^(n-1)+C_3^(n-1)+..+C_(n-1)^(n-1)`

= `1/(0!)+(n-1)/(1!)+((n-1)(n-2))/(2!)+((n-1)(n-2)(n-3))/(3!)+........`

Hence `1/(0!)+(n-1)/(1!)+((n-1)(n-2))/(2!)+((n-1)(n-2)(n-3))/(3!)+........`

= `2^(n-1)`