Question

The least value of
$(x+100{)}^2+(x+99{)}^2+\cdots \cdots +(x+1{)}^2+x^2+(x-1{)}^2+(x-2{)}^2+\dots .+(x-100{)}^2$ is

• A. $6767$
• B. $67670$
• C. $676700$
• D. $767600$

Answer 1

The correct option is C $676700$

$f\left(x\right)=(x+100{)}^2+(x+99{)}^2+\cdots \cdots +(x+1{)}^2+x^2+(x-1{)}^2+(x-2{)}^2+....+(x-100{)}^2$

$f^{{}^{\prime }}\left(x\right)=2(x+100)+2(x+99)+\cdots \cdots +2(x+1)+2x+2(x-1)+2(x-2)+....+2(x-100)=0$

$\Rightarrow x=0$

Since, $f^{{}^{\prime \prime }}\left(0\right)>0$

Therefore, $f\left(x\right)$ is minimum at $x=0$

Therefore, $f\left(0\right)=2(1+2^2+3^2+...+{100}^2)=\frac{2\times 100\times (100+1)\times \left(\right(2\times 100)+1)}{6}=676700$

where sum of square of first $n$ numbers$=\frac{n(n+1)(2n+1)}{6}$