Try this beautiful Problem based on Value of Sum from PRMO 2018, Question 16.
What is the value of $\sum_{1 \leq i<j \leq 10 \atop i+j=\text { odd }}(i-j)-\sum_{1 \leq i<j \leq 10 \atop i+j=\text { even }}(i-j) ?$
Odd-Even
Sum
integer
But try the problem first...
Answer:$55$
PRMO-2018, Problem 16
Pre College Mathematics
First hint
We have to find out the sum . Now substitite $i=1,2,3...9$ and observe the all odd-even cases......
Can you now finish the problem ..........
Second Hint
$i=1 \Rightarrow$$ 1+(2+4+6+8+10-3-5-7-9)$
$=1-4+10=7$
$i=2 \Rightarrow $$0 \times 2+(3+5+7+9-4-6-8-10)$
$=-4$
$i=3 \Rightarrow$$ 1 \times 3+(4+6+8+10-5-7-9)$
$=3-3+10=10$
$i=4 \Rightarrow$$ 0 \times 4+(5+7+9-6-8-10)=-3$
$i=5 \Rightarrow $$1 \times 5+(6+8+10-7-9)=5-2+10$
$=13$
$i=6 \Rightarrow$$ 0 \times 6+(7+9-8-10)=-2$
$i=7 \Rightarrow $$1 \times 7+(8+10-9)=7-1+10=16$
$i=8 \Rightarrow$$ 0 \times 8+(9-10)=-1$
$i=9 \Rightarrow$$ 1 \times 9+(10)=19$
Can you finish the problem........
Final Step
Therefore $ S =(7+10+13+16+19)$-$(4-3-2-1)$ =$55 $
Try this beautiful Problem based on Value of Sum from PRMO 2018, Question 16.
What is the value of $\sum_{1 \leq i<j \leq 10 \atop i+j=\text { odd }}(i-j)-\sum_{1 \leq i<j \leq 10 \atop i+j=\text { even }}(i-j) ?$
Odd-Even
Sum
integer
But try the problem first...
Answer:$55$
PRMO-2018, Problem 16
Pre College Mathematics
First hint
We have to find out the sum . Now substitite $i=1,2,3...9$ and observe the all odd-even cases......
Can you now finish the problem ..........
Second Hint
$i=1 \Rightarrow$$ 1+(2+4+6+8+10-3-5-7-9)$
$=1-4+10=7$
$i=2 \Rightarrow $$0 \times 2+(3+5+7+9-4-6-8-10)$
$=-4$
$i=3 \Rightarrow$$ 1 \times 3+(4+6+8+10-5-7-9)$
$=3-3+10=10$
$i=4 \Rightarrow$$ 0 \times 4+(5+7+9-6-8-10)=-3$
$i=5 \Rightarrow $$1 \times 5+(6+8+10-7-9)=5-2+10$
$=13$
$i=6 \Rightarrow$$ 0 \times 6+(7+9-8-10)=-2$
$i=7 \Rightarrow $$1 \times 7+(8+10-9)=7-1+10=16$
$i=8 \Rightarrow$$ 0 \times 8+(9-10)=-1$
$i=9 \Rightarrow$$ 1 \times 9+(10)=19$
Can you finish the problem........
Final Step
Therefore $ S =(7+10+13+16+19)$-$(4-3-2-1)$ =$55 $
