Proof by induction base case incorrect?

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 $True$

• If you assume $x_n = 1 + \dfrac{1}{2^n}$ is true, can you prove $x_{n + 1} = 1 + \dfrac{1}{2^{n + 1}}$? Apr 5 '18 at 20:52

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?

• I know I may sound ridiculous but to clarify, $x_2$ you would sub in 2? Apr 5 '18 at 20:55
• @KyleMcCann: Is it clear? Apr 5 '18 at 21:00
• Yes that makes more sense, currently trying to solve the rest I will update in a few moments with my answer.. Apr 5 '18 at 21:03
• Slightly confused, I am trying to follow notes (it shows it in 4 steps...) currently on step 2 which is to assume $n = k$ so far I have $x_1=\frac32 x_2=\frac76 x_3=\frac56 x_4=\frac12 + x_k=1+\frac {1}{ 2^k} = notsurewhatgoes here..$ I am just wondering what would I add after $x_k$ Apr 5 '18 at 21:16