3 , 2+1, 1+1+1

The standard proof is as follows. For a given positive integer $n$ , write out a string of $n 1$ 's with small circles between the $1$ 's as follows:

$1 o 1 o 1 o 1.... 1 o 1 o 1$ .

Since there are $n 1$ 's there will be $n - 1$ small circles. We then replace a combination of the circles with $+$ signs to create a particular composition of the integer $n$ .

So, for example, with $n = 3$ , we start with $1 o 1 o 1$ . By replacing only the first circle with a $+$ sign we create the composition $1 + 2$ , and by replacing only the second circle with a $+$ sign we create the composition $2 + 1$ . Replacing both circles gives us $1 + 1 + 1$ , and replacing neither gives us the composition $3$ .

In this way, for any $n$ , we can create a one-to-one correspondence between the sequences as described in the above process and the distinct compositions of $n$ . Since we have $n - 1$ circles we can potentially convert to $+$ signs we have $2^{n-1}$ such sequences, indicating that $a = 2^{n-1}$ , and thus $2a = \boxed{2^n}$ .

• 1 can be rep in 1 way(1)

• 2 can be rep in 2 ways( 2,1+1)

• also 3 can be rep in 4 ways(as given)

• then,4 can be rep in 8 ways and so on........and above follows a series::>2^(n-1)

• hence if n can be rep in 'a' ways i.e. 2^(n-1)=a

then,..the value of 2a=>2*2^(n-1)=2^n

Can done easily by putting value of n in all options If anyone have any considerable soln plz let me know