MQ29 --- Series

∑(b = 0 to a) 2^b
= [2^(a+1) - 1]/(2-1) [by sum of G.S.]
= 2^(a+1) - 1
Hence ∑(a = 0 to n)∑(b = 0 to a) 2^b
= ∑(a = 0 to n) [2^(a+1) - 1]
= ∑(a = 0 to n) [2^(a+1)] - ∑(a = 0 to n) 1
= 2[2^(n+1) - 1]/(2-1) - n
= 2^(n+2) - 3 - n


2012-07-08 14:29:54 補充：
Last two lines:
= 2[2^(n+1) - 1]/(2-1) - n - 1
= 2^(n+2) - 3 - n