Tính nhanh
a)3+8+13+18 ... +2008+2013
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A= - 2014 +2014 +1 . (-1) +(- 8) - 12/25 - 17/37 +13/25 - 20/37
A = ( -2014+2014) + ( - 1) +(-8) - (12/25 - 13/25 ) - (17/37 +20/37)
A = 0 +(-1) + (-8) +1/25 -1
A =(-9) +0,04 - 1
A = - 9,96
Ta có: \(A=1-2+3-4+5-6+7-8+9\)
\(=(1+9)-(2+8)+(3+7)-(4+6)+5\)
\(=10-10+10-10+5\)
\(=5\)
Vậy \(A=5\)
B = 12 - 14 + 16 - 18 + ... + 2008 - 2010
B = -2 + (-2)+ (-2)+ (-2) + ...+ (-2)
B = -2 . 100
B = -200
\(A=\frac{\left[\left(25-1\right):1+1\right]\left(25+1\right)}{2}=325.\)
\(B=\frac{\left[\left(51-3\right):2+1\right]\left(51+3\right)}{2}=675\)
\(C=\frac{\left[\left(81-1\right):4+1\right]\left(81+1\right)}{2}=861\)
\(=18x\left(\frac{19}{21}+\frac{8}{9}\right)\)
\(=18x\frac{113}{63}\)
\(=\frac{226}{7}\)
\(\frac{3}{2}+\frac{3}{8}+\frac{3}{32}+\frac{3}{128}+\frac{3}{512}\)
\(=\left(\frac{12}{8}+\frac{3}{8}\right)+\left(\frac{12}{128}+\frac{3}{128}\right)+\frac{3}{512}\)
\(=\frac{15}{8}+\frac{15}{128}+\frac{3}{512}\)
\(=\frac{240}{128}+\frac{15}{128}+\frac{3}{512}\)
\(=\frac{255}{128}+\frac{3}{512}\)
\(=\frac{1020}{512}+\frac{3}{512}\)
\(=\frac{1023}{512}\)
\(\frac{3}{2}+\frac{3}{8}+\frac{3}{32}+\frac{3}{128}+\frac{3}{512}=\frac{3}{1.2}+\frac{3}{2.4}+\frac{3}{4.8}+\frac{3}{8.16}+\frac{3}{16.32}\)
\(=\frac{3}{1}-\frac{3}{2}+\frac{3}{2}-\frac{3}{4}+\frac{3}{8}-\frac{3}{8}+\frac{3}{16}-\frac{3}{16}+\frac{3}{32}\)
\(=3+\frac{3}{32}=\frac{3.32}{32}+\frac{3}{32}=\frac{96+3}{32}=\frac{99}{32}\)
cau b ta co9;(x+x+.....+x)-(1+2+....+100)=4950
100x - 5050 =4950
100x =4950+5050
100x =10000
x=10000 ;100
x=100
a) 3 + 8 + 13 + 18 + ... + 2008 + 2013
Áp dụng công thức tính dãy số ta có :
\(3+8+...+2013=\frac{\left[\left(2013-3\right):5+1\right].\left(2013+3\right)}{2}=\frac{403.2016}{2}=403.1008=406224\)
3 + 8 + 13 + 18 +... + 2008 +2013 ( 403 số hạng )
= ( 3 + 2013 ) x 403 : 2
=406224