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) ?

1 Answer
May 24, 2017

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)