$\{v_1,,...,v_{2014}\}$ are linearly independent. Find $\lambda$ so that $\{v_1+v_2,...,v_{2014}+\lambda v_1\}$ are also linearly independent. Problem: Vectors $\{v_1,v_2,...,v_{2014}\}$ are linearly independent. Find $\lambda \in \mathbb R$ so that vectors $\{v_1+v_2,v_2+v_3,...,v_{2014}+\lambda v_1\}$ are also linearly independent. Explain.
Edit (to explain a little bit more): I have to find $\lambda$ so that from linear independent vectors $\{v_1,v_2,...,v_{2014}\}$, I have linear independent vectors  $\{v_1+v_2,v_2+v_3,...,v_{2014}+\lambda v_1\}$. 
I am unsure about my result. 
This is what I know:
Since vectors $\{v_1,v_2,...,v_{2014}\}$ are linearly independent, there are some scalars $\alpha_1, \alpha_2, ..., \alpha_{2014} \in \mathbb R$ where $$\alpha_1 v_1+ \alpha_2 v_2+...+\alpha_{2014}v_{2014}=\overrightarrow 0$$ is possible only if $\alpha_1= \alpha_2= ...= \alpha_{2014}=0$.
Now, we take vectors $\{v_1+v_2,v_2+v_3,...,v_{2014}+\lambda v_1\}$ and some scalars $\beta_1,\beta_2,...,\beta_{2014} \in \mathbb R$ so that:
$$\beta_1(v_1+v_2)+\beta_2(v_2+v_3)+...+\beta_{2014}(v_{2014}+\lambda v_1)=\overrightarrow 0$$
Now we have:
$$(\beta_1+\beta_{2014}\lambda)v_1+(\beta_1+\beta_2)v_2+...+(\beta_{2013}+\beta_{2014})v_{2014}=\overrightarrow 0$$
Since we know that vectors $\{v_1,v_2,...,v_{2014}\}$ are linearly indepedent, we conclude that $$(\beta_1+\beta_{2014}\lambda)=(\beta_1+\beta_2)=...=(\beta_{2013}+\beta_{2014})=0$$
We have:
$$\beta_1=-\beta_{2014}\lambda$$
$$\beta_2=-\beta_1=\beta_{2014}\lambda$$
$$\beta_3=-\beta_2=-\beta_{2014}\lambda$$
$$...$$
$$\beta_{2014}=\beta_{2014}\lambda$$
So, $\lambda=0$ or $\lambda=1$. But, since my vectors are linearly independent, I can get, for example, that some scalar $\alpha$ equals $\lambda \beta$. So, $\lambda$ would have to be equal to zero.
I am not sure about this. 
Thank you for your time.
 A: Let us refer to the new basis as $b_1,b_2,\dots,b_{2014}$, that is to say, $b_1=v_1+v_2$, $b_2=v_2+v_3$, on up to $b_{2014}=v_{2014}+\lambda v_1$
Notice that $b_1-b_2+b_3-b_4+\dots - b_{2014}$ simplifies very nicely.

 $(v_1+v_2)-(v_2+v_3)+(v_3+v_4)-\dots = v_1 + (v_2-v_2)+(v_3-v_3)+(v_4-v_4)+\dots = v_1 - \lambda v_1$

We see then that:

 if $\lambda = 1$ we have $b_1-b_2+\dots-b_{2014}=0$ implying that the set is in fact linearly dependent.

On the other hand, if $\lambda$ is anything else, the proof you give above is very good and close to complete.

 Suppose $\lambda\neq 1$.  Then supposing $\beta_1 b_1 + \beta_2 b_2+\dots + \beta_{2014}b_{2014}=0$, and rewriting this in terms of $v_1,v_2,\dots$, the linear independence of $v_1,v_2,\dots$ implies that $\beta_1=-\beta_2=\beta_3=\dots$ and in particular that $\beta_{2014}=\lambda\beta_{2014}$, but since $\lambda\neq 1$, this implies $\beta_{2014}=0$ further implying that every $\beta_i=0$, hence they are linearly independent.


This reaches the conclusion:

 The vectors are linearly independent if and only if $\lambda\neq 1$.

A: The vectors $\{v_1+v_2,...v_{2014}+\lambda v_1\}$ are linearly independent, iff and only the matrix $A$ whose entries $a_{ij}$ are the coefficients of the $i-$th vector respect the $j-$th vector of the basis $\{v_1,...,v_{2014}\}$ is invertible.
The matrix is
$$
\begin{pmatrix}
1&0&0&0&0&\cdots & \lambda\\
1&1&0&0&0&\cdots & 0\\
0&1&1&0&0& \cdots &0\\
0&0&1&1&0& \cdots &0\\
\cdots & & & & & \cdots & \cdots\\
0&0&0&0&0&1&0\\
0 &0 &0&0&0&1&1
\end{pmatrix}.
$$
Now a simple computation show that this matrix is invertible iff $\lambda\neq1$.
