tính giá trị biểu thức sau
C = 1.3+2.4+3.5+......+49.51
làm nhanh giúp mik với
K
Khách
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NL
Nguyễn Lê Phước Thịnh
CTVHS
12 tháng 4 2023
b: 6B=2*4*6+4*6*6+6*8*6+...+46*48*6+48*50*6
=2*4*6-2*4*6+4*6*8-4*6*8+...-44*46*48+46*48*50-46*48*50+48*50*52
=48*50*52
=>B=20800
d: 9D=1*4*9+4*7*9+...+46*49*9
=1*4*2+1*4*7-1*4*7+1*7*10-1*7*10+...+46*49*52-46*49*43
=1*2*4+46*49*52
=117216
=>D=13024
a: 
CG
0
S
7 tháng 10 2016
Đặt \(A=\frac{1}{1.3}+\frac{1}{2.4}+...+\frac{1}{8.10}\)
\(2A=\frac{2}{1.3}+\frac{2}{2.4}+...+\frac{2}{8.10}\)
\(2A=1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{10}\)
\(2A=1-\frac{1}{10}\)
\(2A=\frac{9}{10}\)
\(A=\frac{9}{10}:2=\frac{9}{20}\)
7 tháng 10 2016
=\(\frac{1}{2}\left(\frac{2}{1.3}+...+\frac{2}{8.10}\right)\)
=\(\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}...+\frac{1}{8}-\frac{1}{10}\right)\)
( chắc chắn có số trái dấu ở phía sau, nên còn lại như sau)
=\(\frac{1}{2}\left(1-\frac{1}{10}\right)=\frac{1}{2}.\frac{9}{10}=\frac{9}{20}\)
CN
1
11 tháng 6 2017
\(\frac{1.3}{2^2}.\frac{2.4}{3^2}.\frac{3.5}{4^2}...\frac{98.100}{99^2}\)
\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{98.100}{99.99}\)
\(=\frac{1.2.3...98}{2.3.4...99}.\frac{3.4.5...100}{2.3.4...99}\)
\(=\frac{1}{99}.\frac{100}{2}\)
\(=\frac{1}{99}.50=\frac{50}{99}\)
24 tháng 3 2019
\(\Leftrightarrow N=\frac{\left(2.3.4....50\right)\left(2.3.4...........50\right)}{\left(1.2.3.........49\right)\left(3.4.5...........51\right)}=\frac{50.2}{51}=\frac{100}{51}\)
25 tháng 3 2019
\(\frac{2^2}{1.3}+\frac{3^2}{2.4}+\frac{4^2}{3.5}+....+\frac{50^2}{49.51}\)
\(=\frac{2^2-1}{1.3}+\frac{3^2-1}{2.4}+....+\frac{50^2-1}{49.51}+\frac{1}{1.3}+\frac{1}{2.4}+....+\frac{1}{49.51}\)
\(=\frac{1}{2}.\left(1+1+...+1\right)+\frac{1}{1}-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{49}-\frac{1}{51}\)
Tự làm tiếp :))
25 tháng 3 2019
tớ nhầm đoạn này tí :((
\(=\left(1+1+....+1\right)+\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{51}\right)\)(49 chữ số 1)
\(=49+\frac{1}{2}.\left[\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{49}\right)-\left(\frac{1}{3}+\frac{1}{4}+...+\frac{1}{51}\right)\right]\)
\(=49+\left(\frac{3}{2}-\frac{1}{50}-\frac{1}{51}\right):2\)Tự tính
25 tháng 5 2022
\(A=\dfrac{1}{2}.\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)\left(1+\dfrac{1}{3.5}\right)....\left(\dfrac{1}{2015.2017}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{2}{1}.\dfrac{2}{3}\right).\left(\dfrac{3}{2}.\dfrac{3}{4}\right).\left(\dfrac{4}{3}.\dfrac{4}{5}\right)....\left(\dfrac{2016}{2015}.\dfrac{2016}{2017}\right)\)
\(=\dfrac{1}{2}.\left(\dfrac{2}{1}.\dfrac{2}{3}\right).\left(\dfrac{3}{2}.\dfrac{3}{4}\right).\left(\dfrac{4}{3}.\dfrac{4}{5}\right).....\left(\dfrac{2016}{2015}.\dfrac{2016}{2017}\right)\)
\(=\dfrac{2016}{2017}\)
Ta có: \(C=1\cdot3+2\cdot4+3\cdot5+\cdots+49\cdot51\)
\(=1\left(1+2\right)+2\left(2+2\right)+\cdots+49\left(49+2\right)\)
\(=\left(1^2+2^2+\cdots+49^2\right)+2\left(1+2+\cdots+49\right)\)
\(=\frac{49\left(49+1\right)\left(2\cdot49+1\right)}{6}+2\cdot\frac{49\cdot50}{2}=\frac{49\cdot50\cdot99}{6}+49\cdot50\)
\(=49\cdot25\cdot33+49\cdot50=49\cdot25\left(33+2\right)=49\cdot25\cdot35=42875\)
Ta có: \(C = 1 \cdot 3 + 2 \cdot 4 + 3 \cdot 5 + \hdots + 49 \cdot 51\)
\(= 1 \left(\right. 1 + 2 \left.\right) + 2 \left(\right. 2 + 2 \left.\right) + \hdots + 49 \left(\right. 49 + 2 \left.\right)\)
\(= \left(\right. 1^{2} + 2^{2} + \hdots + 4 9^{2} \left.\right) + 2 \left(\right. 1 + 2 + \hdots + 49 \left.\right)\)
\(= \frac{49 \left(\right. 49 + 1 \left.\right) \left(\right. 2 \cdot 49 + 1 \left.\right)}{6} + 2 \cdot \frac{49 \cdot 50}{2} = \frac{49 \cdot 50 \cdot 99}{6} + 49 \cdot 50\)
\(= 49 \cdot 25 \cdot 33 + 49 \cdot 50 = 49 \cdot 25 \left(\right. 33 + 2 \left.\right) = 49 \cdot 25 \cdot 35 = 42875\)