If #a_ { n } = 32\cdot ( \frac { 1} { 2} ) ^ { n - 1}#, what is #a_{9}#?

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Answer 1 · 2018-04-05T23:28:55

#a_9=1/8#

We have:

#a_n=32(1/2)^(n-1)#

For whatever #n# is, we just plug that value into the expression.

For #a_9#, #n=9#. Therefore:

#a_9=32*(1/2)^(9-1)#

#=>a_9=32*(1/2)^(8)#

#=>a_9=2^5*1^8/(2^8)#

#=>a_9=2^5*1/(2^8)#

#=>a_9=2^5*(2^-8)#

#=>a_9=2^(5+(-8))#

#=>a_9=2^(-3)#

#=>a_9=1/2^3#

#=>a_9=1/8#