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Sagot :
hello
to find the powers of 3 which are 1 single digit, let's check them out
[tex]\begin{gathered} 3^1=3 \\ 3^2=9 \\ 3^3=27 \end{gathered}[/tex]from the calculation above the numbers which are one single digit are only 3 and 9 and their powers are 1 and 2 respectively
question 2
the next seven terms witout getting the power of 3
we'll start first by identifying what type of sequence is this and proceed accordingly to solve
first term = 3
second term = 9
this is a geometric sequence
to find the 3rd, 4th, 5th....7th term, let's use the formula for gp
[tex]T_n=ar^{n-1}[/tex]for 3rd term
a = first term
r = common ratio = 9/3 = 3
[tex]\begin{gathered} T_3=ar^{3-1} \\ T_3=3\times3^2 \\ T_3=27 \end{gathered}[/tex][tex]\begin{gathered} T_4=ar^3 \\ T_4=3\times3^3 \\ T_4=81 \end{gathered}[/tex][tex]\begin{gathered} T_5=ar^4 \\ T_5=3\times3^4 \\ T_5=243 \end{gathered}[/tex][tex]\begin{gathered} T_6=ar^5 \\ T_6=3\times3^5 \\ T_6=729 \end{gathered}[/tex][tex]\begin{gathered} T_7=ar^6 \\ T_7=3\times3^6 \\ T_7=2187 \end{gathered}[/tex][tex]\begin{gathered} T_8=ar^7 \\ T_8=3\times3^7 \\ T_8=6561 \end{gathered}[/tex][tex]\begin{gathered} T_9=ar^8 \\ T_9=3\times3^8 \\ T_9=19683 \end{gathered}[/tex]from the calculation, the next seve terms are 81, 243, 729, 2187, 6561 and 19683
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