## Task B-1.7.

Calculate $\frac{2015 A-2015 B}{1008}$ if

$$
A=\frac{1^{2}}{1}+\frac{2^{2}}{3}+\frac{3^{2}}{5}+\cdots+\frac{1007^{2}}{2013}, \quad B=\frac{1^{2}}{3}+\frac{2^{2}}{5}+\frac{3^{2}}{7}+\cdots+\frac{1007^{2}}{2015}
$$
Solution.

First, let's determine the difference between the numbers \( A \) and \( B \):

\[
\begin{aligned}
& A-B=\frac{1^{2}}{1}+\frac{2^{2}}{3}+\frac{3^{2}}{5}+\cdots+\frac{1007^{2}}{2013}-\left(\frac{1^{2}}{3}+\frac{2^{2}}{5}+\frac{3^{2}}{7}+\cdots+\frac{1007^{2}}{2015}\right)= \\
& =\frac{1^{2}}{1}+\frac{2^{2}}{3}-\frac{1^{2}}{3}+\frac{3^{2}}{5}-\frac{2^{2}}{5}+\cdots+\frac{k^{2}}{2 k-1}-\frac{(k-1)^{2}}{2 k-1}+\cdots+\frac{1007^{2}}{2013}-\frac{1006^{2}}{2013}-\frac{1007^{2}}{2015}=\quad 2 \text{ points } \\
& =\frac{1^{2}}{1}+\frac{(2-1)(2+1)}{3}+\cdots+\frac{(k-k+1)(k+k-1)}{2 k-1}+\cdots+\frac{(1007-1006)(1007+1006)}{2013}-\frac{1007^{2}}{2015}= \\
& 2 \text{ points } \\
& =\underbrace{1+1+1+\cdots+1}_{1007 \cdot 1}-\frac{1007^{2}}{2015}= \\
& =1007-\frac{1007^{2}}{2015}=\frac{2015 \cdot 1007-1007^{2}}{2015}= \\
& =\frac{1007(2015-1007)}{2015}=\frac{1007 \cdot 1008}{2015}
\end{aligned}
\]

Then,

\[
\frac{2015 A-2015 B}{1008}=\frac{2015(A-B)}{1008}=\frac{2015}{1008} \cdot \frac{1007 \cdot 1008}{2015}=1007
\]

2 points

## COUNTY COMPETITION IN MATHEMATICS

2nd grade - high school - B variant
March 3, 2015.

## IF A STUDENT HAS A DIFFERENT APPROACH TO SOLVING THE TASK, THE COMMITTEE IS OBLIGED TO GRADE AND EVALUATE THAT APPROACH APPROPRIATELY.