For a field $F$ let $p(F)=p$ if the characteristic of $F$ is a prime $p$, and $p(F)=+\infty$ if $F$ is of characteristic zero.

In 2007 I considered the linear extension of the Erdos-Heilbronn conjecture, and conjectured (cf. Sun and Zhao - Linear extension of the Erdos-Heilbronn conjecture) that, for any nonzero elements $\lambda_1,\dotsc,\lambda_n$ of a field $F$ with $p(F)\ne n+1$ and a finite subset $A$ of $F$ we have \begin{multline*} \lvert\{\lambda_1a_1+\ldots+\lambda_n a_n:\ \text{$a_1,\dotsc,a_n$ are distinct elements of $A$}\}\rvert \\\ge\min\{p(F)-\delta,\, n(|A|-n)+1\}, \end{multline*} where $\delta=1$ if $n=2$ and $\lambda_1+\lambda_2=0$, and $\delta=0$ otherwise.

Motivated by the above as well as Question 316142 of mine, here I ask the following question.

QUESTION: Is my following conjecture true?

**Conjecture**. Let $\lambda_1,\ldots,\lambda_n\ (n\ge3)$ be positive integers with $\gcd(\lambda_1,\ldots,\lambda_n)=1$ and $\lambda_{k+1}-\lambda_k\in\{0,1\}$ for all $k=1,\ldots,n-1$. Let $F$ be a field with $p(F)>n+1$. Then, for any finite subset $A$ of $F$ with $|A|\ge n+\delta_{n,3}$ we have
\begin{multline*}
\biggl\lvert\biggl\{\sum_{k=1}^n\lambda_ka_k:\ \text{$a_1,\dotsc,a_n$ are distinct elements of $A$}\biggr\}\biggr\rvert
\\\ge\min\biggl\{p(F),\ (\lambda_1+\ldots+\lambda_n)(|A|-n)+\sum_{k=1}^{\lfloor n/2\rfloor}(n+1-2k)(\lambda_{n+1-k}-\lambda_k)+1\biggr\}.
\end{multline*}

Now let me explain where the lower bound comes from. Suppose that $A$ is just the subset $\{1,\ldots,m\}$ of the rational field $\mathbb Q$. For the set $$S=\{\lambda_1a_1+\ldots+\lambda_na_n:\ \text{$a_1,\dotsc,a_n$ are distinct elements of $A$}\},$$ its minimal element should be $\sum_{k=1}^n\lambda_k(n+1-k)$, while its maximal element should be $\sum_{k=1}^n\lambda_k(m-n+k)$. Note that \begin{multline*} \biggl\lvert\biggl\{\sum_{k=1}^n\lambda_k(n+1-k),\ \dotsc,\ \sum_{k=1}^n\lambda_k(m-n+k)\biggr\}\biggr\rvert \\=(\lambda_1+\ldots+\lambda_n)(m-n)+\sum_{k=1}^{\lfloor n/2\rfloor}(n+1-2k)(\lambda_{n+1-k}-\lambda_k)+1. \end{multline*} If $\lambda_k=k$ for all $k=1,\dotsc,n$, then $$\sum_{k=1}^{\lfloor n/2\rfloor}(n+1-2k)(\lambda_{n+1-k}-\lambda_k)=\sum_{k=1}^{\lfloor n/2\rfloor}(n+1-2k)^2=\frac{n(n^2-1)}6.$$

Any comments are welcome!