Question

Enumerate all the numbers in the power of 3 which are in the ones digit and then find the next 7 terms in the sequence without getting the power of 3.
3^4
3^5
3^6
3^7
3^8
3^9
3^10
3^11
3^12
3^13
3^14
3^15?

Answer 1

Answer:

it is a very long Q pls ask sone

Answer 2

The next 7 term in the sequence are 43046721,129140163,387420489,1162261467,3486784401,1046035203,31381059609.

Step-by-step explanation:

$\[{{3}^{4}},{{3}^{5}},{{3}^{6}},{{3}^{7}},{{3}^{8}},{{3}^{9}},{{3}^{10}},{{3}^{11}},{{3}^{12}},{{3}^{13}},{{3}^{14}},{{3}^{15}},\]$ is the sequence.

Next 7 terms in the sequence

3 to the power of 16 is

$3^{16}= 43046721$

3 to the power of 17 is

$3^{17}= 129140163$

3 to the power of 18 is

$3^{18}= 387420489$

3 to the power of 19 is

$3^{19}= 1162261467$

3 to the power of 20 is

$3^{20}= 3486784401$

3 to the power of 21 is

$3^{21}= 1046035203$

3 to the power of 22 is

$3^{22}= 31381059609$

Hence, The next 7 terms in the sequence is 43046721,129140163,387420489,1162261467,3486784401,1046035203,31381059609.