What is the value of ∑_ 1 ≤ i < j ≤ 10 i+j= odd (i+j)-∑_ 1 ≤ i < j ≤ 10 i+j= even (i+j) ?

Correct answer: 55 Sum of odd = 3+5+7+9+ +19 _9+ 5+7+ +17 _7+ 7+9+ +15 _5+ 9+11+13 _3+11 Sum of even = 4+6+ +18 _8+ 6+8+ +16 _6+ 8+10+12+14 _4+ 10+12 _2 Difference =(-1-1-1 .8 times )+19+(-1-1-1 .6 times )+17+(-1-1-1-1)+15+(-1-1)+13+11 =-8-6-4-2+19+17+15+13+11=75-20=55

Olympiad workbook

Exam: Olympiad workbook
Subject: IOQM
Chapter: Sequences and Series

Olympiad workbook · IOQM · Sequences and Series

What is the value of ∑_ 1 ≤ i < j ≤ 10 i+j= odd (i+j)-∑_ 1 ≤ i < j...

What is the value of

\sum_{\substack{1 \leq i < j \leq 10 \\ i+j=\text { odd }}}(i+j)-\sum_{\substack{1 \leq i < j \leq 10 \\ i+j=\text { even }}}(i+j)

Correct answer

Step-by-step solution

Sum of odd

=\underbrace{3+5+7+9+\ldots \ldots+19}_9+\underbrace{5+7+\ldots \ldots+17}_7+\underbrace{7+9+\ldots \ldots \ldots+15}_5+\underbrace{9+11+13}_3+11

Sum of even

=\underbrace{4+6+\ldots \ldots+18}_8+\underbrace{6+8+\ldots \ldots+16}_6+\underbrace{8+10+12+14}_4+\underbrace{10+12}_2

\text { Difference }=(-1-1-1 \ldots \ldots \ldots .8 \text { times })+19+(-1-1-1 \ldots \ldots \ldots .6 \text { times })+17+(-1-1-1-1)+15+(-1-1)+13+11

=-8-6-4-2+19+17+15+13+11=75-20=55

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