# Find a formula for the general term `T_(n)`  of the sequence. {2/3, 4/9, 8/27...}

rakesh05 | Certified Educator

The terms of the given sequence are   `2/3,4/9,8/27, ......`

Here we can investigate the numerators and denominators separately.

The numerator sequence is   `2,4,8, ......`

i.e.   `2^1,2^2,2^3,......`

So the general term by this pattern is  =`2^n` .

Now the denominator sequence is    `3,9,27, ........`

i.e.  `3^1,3^2,3^3,.....`

So the general term of this pattern is =`3^n` .

So the general term of the given sequence is  =`2^n/3^n`

or,            `(2/3)^n`

So,    `T_n=(2/3)^n` .

durbanville | Certified Educator

Note the pattern of `2/3; 4/9; 8/27` as follows

`2^1/3^1; 2^2/3^2; 2^3/3^3`

The first term (c) or  a = `2/3`

Note that when n=1 the exponent is 1; when n=2, the exponent is 2 and so on

`therefore T_(n)= (2/3)^n`

The general formula is: `T_(n) = (2/3)^n`

oldnick | Student

Is showat once  the ratio   `T_(n+1)/T_n=2/3=cost`

so i'ts a geometrical sequence:

`T_(n+k)=(2/3)^k T_n`

Thus:

`T_n=(2/3)^(n-1)T_1=(2/3)^(n-1) (2/3)=(2/3)^n`

pramodpandey | Student

Let sequence is denoted by `c_n`

`c_1=(2/3)^1`

`c_2=4/9=(2/3)^2`

`c_3=8/27=(2/3)^3`

`...`

`..`

`...`

`c_n=(2/3)^n`