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Answer 1 · 2017-02-28T15:25:57

#a_x=3^(x)#

#color(red)("Corrected logic")#

Let the count number be #i#
Let the ith value in the sequence by #a_i#

The sequence for the end of each week:

#"start of first week "-> [1]#

Then we have sequential end of weeks

#[1xx3=3]"; "[3xx3=9]"; "[3xx9=27].....#

#i=1->a_1=1xx3^1=3#
#i=1->a_2=1xx3^2=9#
#i=3->a_3=1xx3^3=27#

and so on

So for any #i# we have #a_i=3^(i)#

The question uses #x# for any term count so we have

#a_x=3^(x)#

Answer 2 · 2017-03-01T12:37:17

The number of flowers after #x# weeks is given by: #3^x#

We can show the numbers of flowers as a sequence first, with each term being multiplied by 3 to get to the next:

#" "1," "3," "9," "27," "81," "243 ...#

We should recognise that these are the powers of 3.

Compare this with the number of weeks and powers of 3:

Weeks: #" "0," "1," "2," "3," "4," "5 ...#

Powers:#" "3^0," "3^1," "3^2," "3^3," "3^4," "3^5 ...#
Flowers:#" "1," "3," "9," "27," "81," "243 ...#

Therefore, after #x# weeks, the number of flowers will be #3^x#