Question

(×+1)+(×+2)+(×+3)....(×+99)=5148​

Answer 1

$\displaystyle\bf\\(x+1)+(x+2)+...+(x+99)=5148~\Leftrightarrow~99x+(1+2+...+99)=5148~\Leftrightarrow\\99x+\frac{99\cdot 100}2} =5148~\Leftrightarrow~99x+99\cdot 50 = 5148~\Leftrightarrow~99(x+50)=5148~\Leftrightarrow\\x+50=52~\Leftrightarrow~x=2.\\\\\boxed{\bf GENERALIZARE}~:~ (x+1)+(x+2)+...+(x+n)=y~\Leftrightarrow~nx+ \frac{n(n+1)}{2} =y~\Leftrightarrow~2nx+n(n+1)=2y~\Leftrightarrow~2nx=2y-n(n+1)~\Leftrightarrow~x=\frac{2y-n(n+1)}{2n}.$