The sequence $u_1$, $u_2$, $u_3$,... is defined by $$u_1=2\,,\,\,\,\,\,\,\,\,\,u_{k+1}=2u_k+1$$ $$u_n=3\times2^{n-1}-1$$ Show that $$\sum_{r=1}^nu_r=u_{n+1}-(n+2)$$ Prove that it is true for $n=1$ thus prove that $u_1=2$ $$\sum_{r=1}^nu_r=3\times2^{(1)}-1-((1)+2)=2$$ I guess you could alternatively use the fact that $u_1=2$ and use the equation $u_{k+1}=2u_k+1$ $$\sum_{r=1}^nu_r=2(2)+1-(1+2)=2$$ Now to prove that $$\sum_{r=1}^nu_r=u_{n+1}-(n+2)$$ you have to assume that it is true? Or assume that for any given value of $n$ that it is true, same thing though isn't it? Let $k$ be any given value (any integer greater than or equal to $1$ I assume), so $n=k$. Then by proving it true for $n=k+1$ I would be proving it true for $n=k$, is that correct?
$$\sum_{r=1}^ku_r=u_{k+1}-(k+2)$$ If this is equal to $u_1+u_2+u_3+...+u_k$, then $$\sum_{r=1}^{k+1}u_r=u_{k+1}-(k+2)\color{red}{+u_{k+1}}$$ Since you're just adding the $(k+1)$th term onto the sum, $$u_1+u_2+u_3+...+u_k\color{red}{+u_{k+1}}$$ Now using $$u_{k+1}=2u_k+1,\,\,\,\,\,\,u_n=3\times2^{n-1}-1$$ I can show that $$\sum_{r=1}^{k+1}u_r=2(u_{k+1})-(k+2)$$ $$\implies2(2u_k+1)-(k+2)$$ $$\implies2(2(3\times2^{k-1}-1)+1)-(k+2)$$ $$\implies2(3\times2^k-2+1)-(k+2)$$ $$\implies3\times2^{k+1}-2-(k+2)$$ Now I need to prove that $$3\times2^{k+1}-2-(k+2)=2(u_{k+1})-(k+2)$$ by substituting $u_k=3\times2^{k-1}-1{\implies}u_{k+1}=3\times2^{(k+1)-1}-1$ into the RHS $$2(u_{k+1})-(k+2)=2(3\times2^{(k+1)-1}-1)-(k+2)$$ $${\implies}3\times2^{k+1}-2-(k+2)$$
So it does work however part of me feels that this is not the right way to go about proving it. Also one question I have is about the very breif answer given in the mark scheme.
$$\sum_{r=1}^{n}u_r=\sum_{r=1}^n3{\times}2^{r-1}-n=3(2^n-1)-n$$ Why have they changed $n-1$ for $r-1$ and subtracted $n$? What is the purpose of $u_r$?
OK, your way is correct but it is very jugaado. Explaining the given method: $$u_r=3.2^{r-1}-1$$ Sigma both sides $$ =\sum_{r=1}^n u_r =\sum_{r=1}^n (3.2^{r-1}-1)$$ Sigma of sum$$=\sum_{r=1}^n3.2^{r-1}-\sum_{r=1}^n 1 $$ Sigma of product$$ =3.\sum_{r=1}^n 2^{r-1}-\sum_{r=1}^n1$$ Geometric Series and constant sigma$$3\frac{2^n-1}{2-1}-n\\=3(2^n-1)-n$$