100X100—99X99+98X98+...+2X2—1X1等于多少
问题描述:
100X100—99X99+98X98+...+2X2—1X1等于多少
答
S= 100X100—99X99+98X98+...+2X2—1X1
=∑(2n)-∑(2n-1)
=∑(4n-1)
=4n(n+1)/2 - n
=2n(n+1) - n
= 2x50x(50+1) - 50
=5050
答
将一正一负的两项放在一起,利用平方差公式展开,将会得到一个等差数列,再用等差数列求和公式算。
答
=(100+99)(100-99)+(98+97)(98-97)...+(2+1)(2-1)
然后就是一个等差数列,答案为5050
答
S= 100X100—99X99+98X98+...+2X2—1X1
=∑(2n)²-∑(2n-1)²
=∑(4n-1)
=4n(n+1)/2 - n
=2n(n+1) - n
= 2x50x(50+1) - 50
=5050