Question

lets see who can solve this question-

Answer 1

Given,

$\longrightarrow\sf{x=1+1-1+1-1+1-1+\dots}$

$\longrightarrow\sf{x=1+(1-1)+(1-1)+(1-1)+\dots}$

$\longrightarrow\sf{x=1+0+0+0+\dots}$

$\longrightarrow\sf{x=1\quad\quad\dots(1)}$

Also,

$\begin{array}{cccccccccccccccccc}\longrightarrow&\sf{x}&=&\sf{1}&+&\sf{1}&-&\sf{1}&+&\sf{1}&-&\sf{1}&+&\sf{1}&-&\sf{1}&+&\sf{1}-\dots\\\\+&\sf{\dfrac{x}{2}}&=&&&\sf{\dfrac{1}{2}}&+&\sf{\dfrac{1}{2}}&-&\sf{\dfrac{1}{2}}&+&\sf{\dfrac{1}{2}}&-&\sf{\dfrac{1}{2}}&+&\sf{\dfrac{1}{2}}&-&\sf{\dfrac{1}{2}}+\dots\\\\\cline{1-18}\\&\sf{\dfrac{3x}{2}}&=&\sf{1}&+&\sf{\dfrac{3}{2}}&-&\sf{\dfrac{1}{2}}&+&\sf{\dfrac{1}{2}}&-&\sf{\dfrac{1}{2}}&+&\sf{\dfrac{1}{2}}&-&\sf{\dfrac{1}{2}}&+&\sf{\dfrac{1}{2}}-\dots\end{array}$

$\longrightarrow\sf{\dfrac{3x}{2}=1+\dfrac{3}{2}-\left(\dfrac{1}{2}-\dfrac{1}{2}\right)-\left(\dfrac{1}{2}-\dfrac{1}{2}\right)-\left(\dfrac{1}{2}-\dfrac{1}{2}\right)-\dots}$

$\longrightarrow\sf{\dfrac{3x}{2}=1+\dfrac{3}{2}-0-0-0-\dots}$

$\longrightarrow\sf{\dfrac{3x}{2}=1+\dfrac{3}{2}}$

$\longrightarrow\sf{\dfrac{3x}{2}=1+1+\dfrac{1}{2}}$

From (1),

$\longrightarrow\sf{\dfrac{3x}{2}=x+x+\dfrac{1}{2}}$

$\longrightarrow\sf{\dfrac{3x}{2}=2x+\dfrac{1}{2}}$

$\longrightarrow\sf{\dfrac{3x}{2}-2x=\dfrac{1}{2}}$

$\longrightarrow\sf{-\dfrac{x}{2}=\dfrac{1}{2}}$

$\longrightarrow\sf{x=-1\quad\quad\dots(2)}$

Adding (1) and (2),

$\longrightarrow\sf{x+x=1-1}$

$\longrightarrow\sf{2x=0}$

$\longrightarrow\sf{x=0\quad\quad\dots(3)}$

Adding (1) and (3),

$\longrightarrow\sf{x+x=1+0}$

$\longrightarrow\sf{2x=1}$

$\longrightarrow\sf{\underline{\underline{x=\dfrac{1}{2}}}}$

Hence Proved!