How to prove this by induction?

Question

The sequence ${u_n}$ is given by the recurrence relation

$u_{n+1}=u_n-2 $ for n = 1,2,3, ...,

and the initial term $u_1=-1$

Prove by induction that $\sum^m_{n=1} u_n = -m^2$ for att $m\geq 1$

I get stuck in the induction step.

Answer 1

Clearly the result holds for $m=1:$ $\sum_{n=1}^{1}u_n=u_1=-1=-1^2$.

Assume the result holds for $m=k$ for some $k\in \mathbb{N}$. Then $$\sum_{n=1}^{k}u_n=-k^2.$$

Consider when $n=k+1$. We have $u_k=-2k+1$ and so $$\begin{align} \sum_{n=1}^{k+1}u_n&=u_{k+1}+\sum_{n=1}^{k}u_n \\ &=(u_k-2)-k^2 \\ &=-2k-1-k^2 \\ &=-(k+1)^2. \end{align}$$

Hence the result holds by induction.

Answer 2

First of all show $u_n=-2n+1$ for all $n$

Then you have $$\sum_{n=1}^{m+1} u_n=\sum_{n=1}^m u_n+u_{m+1}=-m^2+u_m-2=-m^2-2m+1-2=-m^2-2m-1=-(m+1)^2$$