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Question
Simplify the following:
`(3^nxx9^(n+1))/(3^(n-1)xx9^(n-1))`
Solution
`(3^nxx9^(n+1))/(3^(n-1)xx9^(n-1))`
`=(3^nxx(3^2)^(n+1))/(3^(n-1)xx(3^2)^(n-1))`
`=(3^nxx3^(2n+2))/(3^(n-1)xx3^(2n-2))`
`=3^(n+2n+2)/3^(n-1+2n-2)`
`=3^(3n + 2)/3^(3n-3)`
`=3^(3n+2-3n+3)`
= 35
= 243
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Find:-
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