Show that $\sum_{r=1}^nu_r=u_{n+1}-(n+2)$

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The sequence $u_1$, $u_2$, $u_3$,... is defined by $$u_1=2\,,\,\,\,\,\,\,\,\,\,u_{k+1}=2u_k+1$$ Then I was asked to prove that, for all $n\ge1$ $$u_n=3\times2^{n-1}-1$$ Now I'm being asked to show that $$\sum_{r=1}^nu_r=u_{n+1}-(n+2)$$ This has left me very confused. I thought $\sum_{r=1}^nu_r$ just meant $u_n$? The sum from $r$ to $n$ where $r=1$. Problem is that $$u_n\,\,{\ne}\,\,u_{n+1}-(n+2)$$ What does $u_r$ represent?

Answer 1

If you already know that $\;u_n=3\cdot2^{n-1}-1\;$ , then carry on the sum explicitly:

$$\sum_{r=1}^nu_r=3\cdot\sum_{r=1}^n2^{r-1}-\sum_{r=1}^n1=3\frac{2^n-1}{2-1}-n=$$

$$=3\cdot2^n-3-n=\left(3\cdot2^{2n-1}-1\right)-2-n=u_{n+1}-(n+2)$$

Answer 2

First, the sum should be read like this: "a sum of elements from 1-st to n-th is equal to (n+1)-st element minus (n+2)". For example, $u_1 + u_2 + u_3 = u_4 - (3+2)$ (you can verify this by calculating the corresponding elements.

I suggest you use the mathematical induction here: for $n = 1$, the sum is equal to one element: $\sum_{r=1}^1 u_r = u_1 = 2$. Now, the right side is easy: $u_{n+1} - (n+2) = u_2 - (1+2) = 2u_1 + 1 - (1+2) = 2*2 + 1 - 3 = 2$

So, we've proved this for $n = 1$.

Now, try to show that if it is true for any given $n$, then it will be true for the $n+1$. This concludes the induction!

Are you having trouble proving that $u_n = 3\times 2^{n-1} - 1$?