What is #(1+(2/1))(1+(2/2))(1+(2/3))...(1+(2/27))#?

1 Answer
George C.
Apr 6, 2018

#406#

Explanation:

#(1+2/1)(1+2/2)(1+2/3)...(1+2/27)#

#=((1+2)/1)((2+2)/2)((3+2)/3)...((27+2)/27)#

#=(color(red)(cancel(color(black)(3)))/1)(color(brown)(cancel(color(black)(4)))/2)(color(orange)(cancel(color(black)(5)))/color(red)(cancel(color(black)(3))))(color(green)(cancel(color(black)(6)))/color(brown)(cancel(color(black)(4))))...(color(violet)(cancel(color(black)(27)))/color(blue)(cancel(color(black)(25))))(28/color(purple)(cancel(color(black)(26))))(29/color(violet)(cancel(color(black)(27))))#

#=(28 * 29)/(1 * 2) = 406#

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