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)# ?

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Answer 1 · 2017-05-24T07:30:57

Please see below.

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)#