Question

99²－1²＋3²－5²＋7²－……＋95²－97²等于多少？

Answer 1

99^2-1^2+3^2-5^2+7^2-…+95^2-97^2
=99^2-97^2+95^2-93^2+…+7^2-5^2+3^2-1^2
=(99+97)(99-97)+(95+93)(95-93)+…+(7+5)(7-5)+(3+1)(3-1)
=2(99+97+95+93+…+7+5+3+1)
=2×[(99+1)/2]^2
=2×50^2
=5000
注:1+3+5+7+…+(2n-1)=n^2(n为正整数)

Answer 2

=(3²-1²)+(7²-5²)+……+(95²-93²)+(99²-97²)
=(3+1)(3-1)+(7+5)(7-5)+……+(95+93)(95-93)+(99+97)(99-97)
=2(1+3)+2(5+7)+……+2(93+95)+2(97+99)
=2(1+3+5+7+……+93+95+97+99)
=2×[(99+1)/2]²
=2×50²
=5000
1+3=2²
1+3+5=3²
1+3+5+7=4²
1+3+5+7+……+(2n-1)=n²

Answer 3

99²－1²＋3²－5²＋7²－……＋95²－97²
=(3²－1²)＋(7²－5²)＋……＋(95²-93²)+(99²－97²)
=(3+1)(3-1)+(7+5)(7-5)+......(95+93)(95-93)+(99+97)(99-97)
=(1+3)×2+(5+7)×2+......+(93+95)×2+(97+99)×2
=(1+3+5+7+......+94+95+97+99)×2
={(1+99)×[(99-1)÷(3-1)+1]}×2
={100×(98÷2+1)}×2
=(100×50)×2
=10000