Use mathematical induction to prove that for all positive integers n, $ \sum_{r=1}^n (2r-1)\cdot 2^{-r} = 3- \frac{2n+3}{2^n} $.

Question

Use mathematical induction to prove that for all positive integers n,

$$ \sum_{r=1}^n (2r-1) \cdot 2^{-r} = 3 - \frac{2n+3}{2^n} $$

I have managed to prove the one above but I am unable to continue the second part.

Hence, show that

$$ \sum_{r=1}^n r \cdot 2^{-r} = 2 - \frac{n+2}{2^n}. $$

Answer 1

$$\sum_{r=1}^nr2^{-1}=\frac12\left(\sum_{r=1}^n(2r-1)2^{-r}+\sum_{r=1}^n2^{-r}\right)$$

So find an expression for $\sum_{r=1}^n2^{-r}$ and substitute.

Answer 2

If $$s_n=\sum_{r=1}^n (2r-1) \cdot 2^{-r}$$ then $$s_{n+1} = s_n+{2n+1\over 2^{n+1}}$$

Does this help?

Answer 3

$S(n):=\sum_{r=1}^n 2^r$ and $\alpha=\frac{1}{2}$

$T(n)=2\sum_{r=1}^{n}r \alpha^r=$

$2\alpha (\sum_{r=2}^{n+1} r\alpha^{r-1}+1-(n+1)\alpha^n)=$

$2\alpha (\sum_{r=2}^{n+1} (r-1)\alpha^{r-1}+1+\sum_{r=2}^{n+1}\alpha^{r-1}-(n+1)\alpha^{n})= $

$=2\alpha(\frac{1}{2}T(n)+1+S(n)-(n+1)\alpha^{n})$

and so

$T(n)=\frac{2\alpha}{1-\alpha}(1-(n+1)\alpha^n+S(n))$

Now

$\sum_{k=1}^n (2r-1)2^{-r}=T(n)-S(n)=$

$=2-2(n+1)\frac{1}{2}^n+S(n)$