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

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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.

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` **

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`

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`