There must be a typo. For k =1, the condition $|a_{1}+2a_{2}+3a_{3}+\cdots+k\cdot a_{k}|\leq 2k+1$ becomes $|a_1| \leq 3$, which is true because $a_1$ lies in $[-1,1]$. Perhaps you mean to prove something else about it.

gauss202 wrote: There must be a typo. For k =1, the condition $|a_{1}+2a_{2}+3a_{3}+\cdots+k\cdot a_{k}|\leq 2k+1$ becomes $|a_1| \leq 3$, which is true because $a_1$ lies in $[-1,1]$. Perhaps you mean to prove something else about it. Sorry, it's been edited.

Does any one have full solution??

Assume that $|a_1+2a_2+\ldots+ka_k|>\frac{2k+1}{4}$ for all $k$ between $1$ and $n$. Suppose WLOG that $a_1>0$ and hence $a_1>\frac{3}{4}$ by assumption. Then if there is some $m$ so that $a_1+2a_2+\ldots+ma_m>\frac{2m+1}{4}$ but $a_1+2a_2+\ldots+(m+1)a_{m+1}<-\frac{2(m+1)+1}{4}$, then we get $(m+1)a_{m+1}<-\frac{2(m+1)+1}{4}-\frac{2m+1}{4}=-(m+1)$ and so $a_{m+1}<-1$, which contradicts our assumptions. Thus $a_1+2a_2+\ldots+ka_k>\frac{2k+1}{4}$ for all $k$ between $1$ and $n$. Now we prove that $\frac{k-1}{k}a_1+\frac{k-2}{k}a_2+\ldots+\frac{1}{k}a_{k-1}>0$ for all $k$ between $2$ and $n$. It's true for 2, because $\frac{1}{2}a_1>0$. If it's true for $k$, then we can add $\frac{1}{k(k+1)}(a_1+2a_2+\ldots+ka_k)>0$ to both sides: the coefficient of $a_i$ is $\frac{k-i}{k}+\frac{i}{k(k+1)}=\frac{k^2+k-ik-i+i}{k(k+1)}=\frac{k+1-i}{k+1}$. Hence $\frac{k}{k+1}a_1+\frac{k-1}{k+1}a_2+\ldots+\frac{1}{k+1}a_k>0$. So by induction, it's true for all $k$ between $2$ and $n$. Finally, we add in $\frac{1}{k}(a_1+2a_2+\ldots+ka_k)>0$ to get $a_1+a_2+\ldots+a_k>0$ for all $k$ from $2$ to $n$, including $a_1+a_2+\ldots+a_n>0$, in contradiction to what we assumed to start.

Problem 4, 2006 Romanian Selection Tests, http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=84392&p=492891#p492891.