\(a_1+a_2+a_3+..+a_{2015}=0\)\(0\)
\(\Rightarrow\left(a_1+a_2\right)+...+\left(a_1+a_{2015}\right)\)\(=\frac{\left(2015-1\right)}{2}+1=1008\)
\(\Rightarrow a_1+\left(a_1+a_2+..+a_{2015}\right)=1008\)
\(\Rightarrow a_1=1008\)
Ta có:
\(a_1+a_2+...+a_{2015}=0\)
\(\Leftrightarrow\left(a_1+a_2\right)+\left(a_3+a_4\right)+...+\left(a_{2013}+a_{2014}\right)+\left(a_{2015}+a_1\right)-a_1=0\)
\(\Leftrightarrow1+1+...+1-a_1=0\)
\(\Leftrightarrow1008-a_1=0\)
\(\Leftrightarrow a_1=1008\)
Theo giả thiết, ta có:
\(a_1+a_2=a_3+a_4=...=a_{2015}+a_1=1\)
\(\Rightarrow\left(a_1+a_2\right)+\left(a_3+a_4\right)+...+\left(a_{2015}+a_1\right)=1.\left[\frac{\left(2015-1\right)}{2}+1\right]=1008\)
\(\Leftrightarrow\left(a_1+a_2+a_3+...+a_{2015}\right)+a_1=1008\)
\(\Leftrightarrow0+a_1=1008\Rightarrow a_1=1008\)
\(a_1+a_2=a_3+a_4=.....=a_{2015}+a_1=1\\ \Rightarrow a_1+a_2+a_3+a_4+....+a_{2015}+a_1=1008\\ \)
\(\Rightarrow0+a_1=2008\\ \Rightarrow a_1=2008\)
Chúc bạn học tốt!!!!!!!
a1+a2=a3+a4=...=a2015+a1=1
\(\Rightarrow\)(a1+a2)+(a3+a4)+...+(a2015+a1)=1.[\(\frac{\left(2015-1\right)}{2}\)+1]=1008
\(\Leftrightarrow\)(a1+a2+a3+...+a2015)+a1=1008
\(\Leftrightarrow\)0+a1=1008\(\Rightarrow\)a1=1008
Ta có a1 + a2 + a3 +... + a 2015 + a1 = 1008
=> 0 + a1 = 2008
=> a1 = 2008
\(\Rightarrow\left(a_1+a_2\right)+\left(a_3+a_4\right)....+a_{2015}=0\)
\(=>1+1+...+a_{2015}=0\)
\(\Rightarrow1007+a_{2015}=0\Rightarrow a_{2015}=-1007\)
mà \(a_1+a_{2015}=1\Rightarrow a_1=1008\)
