The question goes as the following:
A sequence of real numbers $x_1$,$x_2$,..,$x_n$ is defined by the recursion
$\ x_1 = \frac{3}{2}$ and $x_{k+1} = x_k - \frac{1}{(2^k+1)}$ for $k$ $\geqslant$ $1$
Calculate $x_2,x_3$ and $x_4$ and then prove by induction that
$x_n$ = 1 + $\frac{1}{2^n}$ for $n \geqslant 1.$
My attempt so far has been to do the base case which would let $n = 1$ and sub in for $n$.
What I have is the following: $x_1$ = 1 + $\frac{1}{2^1}$ which is the same as $\frac{3}{2}$
Which results in: x1 = 1.5 This would mean that the base case is not True?
I am wondering how would this be solved to obtain $x_2$,$x_3$. Since $x1$ is given and proven $True$
Hint
For obtain $x_2$ use that $x_2=x_1-\frac{1}{2^1+1}=\frac32-\frac13=\frac76$.
Use the same idea for $x_3$ and $x_4$.
By induction, you need assume that it is true for $x_k$ $\left(\text{that means } x_k=1+\frac {1}{ 2^k}\right)$ and prove for $x_{k+1}$ $\left(\text{that means } x_{k+1}=1+\frac {1}{ 2^{k+1}}\right)$, using that $x_{k+1}=x_k-\frac{1}{2^k+1}$.
It means
$$x_{k+1}=1+\frac {1}{ 2^k}-\frac{1}{2^k+1}$$
can you finish?