Thực hiện phép tính:
C=1+3+6+10+15+...+1225
Giải giúp mk nha.
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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
\(=\left(1-2-3\right)+4+\left(5-6-7\right)+8+...+\left(77-78-79\right)+80\)\(=\left(-4\right)+4+\left(-8\right)+8+...+\left(-80\right)+80\)\(=0+0+...+0=0\)\(A=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.....\frac{9999}{10000}\)
\(A=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.....\frac{99.101}{100.100}\)
\(A=\frac{\left(1.2.3.....99\right).\left(3.4.5.....101\right)}{\left(2.3.4.....100\right).\left(2.3.4.....100\right)}\)
\(A=\frac{1.101}{2.100}=\frac{101}{200}\)
\(A=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}......\frac{9999}{10000}\)
\(A=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.....\frac{99.101}{100.100}\)
\(A=\frac{1.2.3.4.....99}{2.3.4.5.....100}.\frac{3.4.5.6.....101}{2.3.4.5.....100}\)
\(A=\frac{1}{100}.\frac{101}{2}\)
\(A=\frac{101}{200}\)
(315x4+5.315):315
=(57395628+71744535):315
=129140163:315
=9
\(\frac{-1}{-16}-\frac{1}{15}\)
\(=\frac{1}{16}-\frac{1}{15}\)
\(=\frac{15-16}{16\times15}\)
\(=\frac{-1}{240}\)
\(-\frac{1}{16}-\frac{1}{15}\)
\(=-\left(\frac{1}{16}+\frac{1}{15}\right)\)
\(=-\frac{16+15}{16\times15}\)
\(=-\frac{31}{240}\)
\(2C=2+6+12+20+...+2450\)
\(2C=1.2+2.3+3.4+4.5+...+49.50\)
\(6C=1.2.3+2.3.3+3.4.3+...+49.50.3\)\(6C=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+49.50.\left(51-48\right)\)
\(6C=1.2.3+2.3.4-1.2.3+...+49.50.51-49.50.51\)
\(6C=49.50.51\)
\(6C=124950\)
\(C=20825\)
số các số hạng là:(1225-1):+1=613 (số hạng)
tổng C =(1225+1)x 613:2=375769