How do you use sigma notation to write the sum for #1/2+2/4+6/8+24/16+120/32+720/64#?

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Answer 1 · 2017-06-19T13:00:22

The answer is #=sum_1^6(n!)/(2^n)#

The numerator is #=n!#

The denominator is #=2^n#

Therefore,

#1/2+2/4+6/8+24/16+120/32+720/64+..........#

#=sum_1^n(n!)/(2^n)#

In this particular case, it is

#=sum_1^6(n!)/(2^n)#

where,

#n! =1*2*3*4...n#