Value of Sum | PRMO - 2018 | Question 16

Try this beautiful Problem based on Value of Sum from PRMO 2018, Question 16.

Value of Sum - PRMO 2018, Question 16

What is the value of $\sum_{1 \leq i<j \leq 10 \atop i+j=\text { odd }}(i-j)-\sum_{1 \leq i<j \leq 10 \atop i+j=\text { even }}(i-j) ?$

$50$
$53$
$55$
$59$
$65$

Key Concepts

Odd-Even

Sum

integer

Check the Answer

Answer: $55$

PRMO-2018, Problem 16

Pre College Mathematics

Try with Hints

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 ..........

$i=1 \Rightarrow <img src="https://www.cheenta.com/wp-content/ql-cache/quicklatex.com-8de8b5ab6f27dc944320b7761c54e597_l3.png" height="69" width="444" class="ql-img-displayed-equation quicklatex-auto-format" alt="\[1+(2+4+6+8+10-3-5-7-9)$$=1-4+10=7$$i=2 \Rightarrow\]" title="Rendered by QuickLaTeX.com"/>0 \times 2+(3+5+7+9-4-6-8-10)$

$=-4$

$i=3 \Rightarrow <img src="https://www.cheenta.com/wp-content/ql-cache/quicklatex.com-6ac051fd0449f99bad8ef6e8ee04c505_l3.png" height="69" width="429" class="ql-img-displayed-equation quicklatex-auto-format" alt="\[1 \times 3+(4+6+8+10-5-7-9)$$=3-3+10=10$$i=4 \Rightarrow\]" title="Rendered by QuickLaTeX.com"/> 0 \times 4+(5+7+9-6-8-10)=-3$

$i=5 \Rightarrow <img src="https://www.cheenta.com/wp-content/ql-cache/quicklatex.com-1d28ea5c5ef82ca63b590c0706a5ba25_l3.png" height="69" width="451" class="ql-img-displayed-equation quicklatex-auto-format" alt="\[1 \times 5+(6+8+10-7-9)=5-2+10$$=13$$i=6 \Rightarrow\]" title="Rendered by QuickLaTeX.com"/> 0 \times 6+(7+9-8-10)=-2$

$i=7 \Rightarrow <img src="https://www.cheenta.com/wp-content/ql-cache/quicklatex.com-d70bb77c7d9ff1ea530a5066730cea1d_l3.png" height="47" width="441" class="ql-img-displayed-equation quicklatex-auto-format" alt="\[1 \times 7+(8+10-9)=7-1+10=16$$i=8 \Rightarrow\]" title="Rendered by QuickLaTeX.com"/> 0 \times 8+(9-10)=-1$

$i=9 \Rightarrow$

$1 \times 9+(10)=19$

Can you finish the problem........

Therefore $S =(7+10+13+16+19)$ - $(4-3-2-1)$ = $55$