Question regarding an induction proof

Question

I am stuck on a question regarding induction. I know that we are supposed to solve it using 3 steps: the base step, the n= p step and n = p+1.

The question is prove that $$\displaystyle\sum_{i=1}^n\dfrac{i}{2^i}= 2- \dfrac{n+2}{2^n}$$

For $n = 1$ both sides will be $\dfrac{1}{2}$.

It is on the step $n+1$ that I am stuck.

I have calculated LHS to:

$$2-\dfrac{n+2}{2^n} + \dfrac{n+1}{2^{n+1}}$$

On the RHS for $n+1$ I have:

$$2-\dfrac{(n+1)+2}{2^{n+1}}$$

I would be thankful for any ideas on how to continue. My guess has to do with finding common factors which could be: $2$, $n+1$ or $2^{n+1}$.

Let me know if anything needs to be clarified.

Answer 1

you must show that $$2-\frac{n+2}{2^n}+\frac{n+1}{2^{n+1}}=2-\frac{n+3}{2^{n+1}}$$

Answer 2

For the Induction Step,

$2-\dfrac{n+2}{2^n}+\dfrac{n+1}{2^{n+1}}=2-\dfrac{1}{2^n}\left(n+2-\dfrac{n+1}{2}\right)=2-\dfrac{1}{2^n}\left(\dfrac{2n+4 -n-1}{2}\right)=2-\dfrac{n+3}{2^{n+1}}$

Answer 3

You forgor some parentheses around the exponents:

Left-hand side: $\quad 2-\dfrac{n+2}{2^n}+\dfrac{n+1}{2^{n+1}}=2-\dfrac{2(n+2)}{2^{n+1}}+\dfrac{n+1}{2^{n+1}}=2-\dfrac{n+3}{2^{n+1}}.$

Right-hand side: $\quad 2-\dfrac{(n+1)+2}{2^{n+1}}$.