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\(C=-\left[\dfrac{1}{3}\cdot\dfrac{\left(3+1\right)\cdot3}{2}+\dfrac{1}{4}\cdot\dfrac{\left(4+1\right)\cdot4}{2}+...+\dfrac{1}{50}\cdot\dfrac{\left(50+1\right)\cdot50}{2}\right]\\ C=-\left(\dfrac{1}{3}\cdot\dfrac{4\cdot3}{2}+\dfrac{1}{4}\cdot\dfrac{5\cdot4}{2}+...+\dfrac{1}{50}\cdot\dfrac{51\cdot50}{2}\right)\\ C=-\left(2+\dfrac{5}{2}+...+\dfrac{51}{2}\right)\\ C=-\dfrac{4+5+...+51}{2}=-\dfrac{\dfrac{\left(51+4\right)\left(51-4+1\right)}{2}}{2}=-\dfrac{55\cdot48}{4}=-660\)
1-2-3+4+5-6-7+8+9-10-11+......................-78+80
\(=\left(1-2-3\right)+4+\left(5-6-7\right)+8+...+\left(77-78-79\right)+80\)\(=\left(-4\right)+4+\left(-8\right)+8+...+\left(-80\right)+80\)\(=0+0+...+0=0\)(315x4+5.315):315
=(57395628+71744535):315
=129140163:315
=9
\(\frac{-1}{-16}-\frac{1}{15}\)
\(=\frac{1}{16}-\frac{1}{15}\)
\(=\frac{15-16}{16\times15}\)
\(=\frac{-1}{240}\)
\(-\frac{1}{16}-\frac{1}{15}\)
\(=-\left(\frac{1}{16}+\frac{1}{15}\right)\)
\(=-\frac{16+15}{16\times15}\)
\(=-\frac{31}{240}\)
\(2C=2+6+12+20+...+2450\)
\(2C=1.2+2.3+3.4+4.5+...+49.50\)
\(6C=1.2.3+2.3.3+3.4.3+...+49.50.3\)\(6C=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+49.50.\left(51-48\right)\)
\(6C=1.2.3+2.3.4-1.2.3+...+49.50.51-49.50.51\)
\(6C=49.50.51\)
\(6C=124950\)
\(C=20825\)
a,
A=1−3−5−7−9−...−97−99a)A=1−3−5−7−9−...−97−99
=1−(3+5+7+...+99)=1−(3+5+7+...+99)
=1−(99+3).[(99−3):2+1]2=1−(99+3).[(99−3):2+1]2
=1−2499=−2498=1−2499=−2498
b)B=1+3−5−7+9+...+97−99b)B=1+3−5−7+9+...+97−99
=(−8)+(−8)+(−8)+...+(−8)+97−99=(−8)+(−8)+(−8)+...+(−8)+97−99
=(−8).12+(−2)=−98=(−8).12+(−2)=−98
c)C=1−3−5+7+9−11−13+15+...+97−99c)C=1−3−5+7+9−11−13+15+...+97−99
=0+0+0+0+0+...+0−99=0+0+0+0+0+...+0−99
=−99
Bn đã đăng câu hỏi này lúc nãy rồi mà ?
Sao bn lại đăng lại ?
số các số hạng là:(1225-1):+1=613 (số hạng)
tổng C =(1225+1)x 613:2=375769