Find a formula for the general term `T_(n)`  of the sequence: {2,4,8,16,32}

4 Answers | Add Yours

rakesh05's profile pic

rakesh05 | High School Teacher | (Level 1) Assistant Educator

Posted on

Given sequence is  `{2,4,8,16, ......}.`

`` Here we can cosider the first term  as  `T_1=2`

                             second term  as    `T_2=4`

`T_3=8` , `T_4=16`

We observe by this pattern that every successive term is twice the predecessor and can be written as powers of 2.

i.e.    `T_1=2^1`

`T_2=2^2` 

`T_3=2^3`   and so on.   So in this way by this pattern we can write

                        `T_n=2^n` .  Answer.

 

durbanville's profile pic

durbanville | High School Teacher | (Level 2) Educator Emeritus

Posted on

2  ;  4  ;  8  ;  16  ;  32

Note that the pattern is `2^1; 2^2; 2^3; 2^4 ;2^5`

`T_(1) = 2^1` `T_(2) = 2^2` and so on

`therefore T_(n) = 2^n`

Note that when n=1, the exponent is 1. When n=2, the exponent is 2 and so on.  (the first term)a = 2

The general formula is therefore `T_(n) = 2^n`

oldnick's profile pic

oldnick | (Level 1) Valedictorian

Posted on

First we have to take note that   `T_(n+1)=2T_n`

So we have `T_(n+2)=2T_(n+1)=4T_n`

That is:      `T_(n+k)=2^kT_n`

so:`T_(n)=2^(n-1)T_1=2^(n-1)2=2^n`

 

pramodpandey's profile pic

pramodpandey | College Teacher | (Level 3) Valedictorian

Posted on

let sqeuence is denoted by  `c_n`

then

`c_1=2`

`c_2=4`

`c_3=8`

`c_4=16`

`c_5=23`

`.`

`.`

`c_n=2^n ,n inN`

We’ve answered 318,915 questions. We can answer yours, too.

Ask a question