1, A= 4+42+43+44+...+499+4100
Chứng minh A \(⋮\)5
Tính:
a, S = 1.2+2.3+3.4+...+2018.2019
b, P = 1.2.3+2.3.4+...+48.49.50
c, Q = 12+22+32+42+...+1132
K
Khách
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Những câu hỏi liên quan
HL
1
14 tháng 6 2020
\(A=\frac{4}{1.2}+\frac{4}{2.3}+\frac{4}{3.4}+...+\frac{4}{2019.2020}\)
\(\frac{1}{4}A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\)
\(\frac{1}{4}A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\)
\(\frac{1}{4}A=1-\frac{1}{2020}=\frac{2019}{2020}\)
\(\Rightarrow A=\frac{2019}{2020}:\frac{1}{4}=\frac{2019}{505}\)
Vậy \(A=\frac{2019}{505}.\)
\(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}\)
\(\Rightarrow2B=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{98.99.100}\)
\(2B=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{98.99}-\frac{1}{99.100}\)
\(2B=\frac{1}{1.2}-\frac{1}{99.100}=\frac{4949}{9900}\)
\(\Rightarrow B=\frac{4949}{9900}:2=\frac{4949}{19800}\)
Vậy \(B=\frac{4949}{19800}.\)
14 tháng 6 2020
\(A=\frac{4}{1\cdot2}+\frac{4}{2\cdot3}+\frac{4}{3\cdot4}+...+\frac{4}{2019\cdot2020}\)
\(A=4\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2018\cdot2019}\right)\)
\(A=4\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2018}-\frac{1}{2019}\right)\)
\(A=4\left(1-\frac{1}{2019}\right)=4\cdot\frac{2018}{2019}\)
Đến đây tự tính
\(B=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{98\cdot99\cdot100}\)
\(B=\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{98\cdot99\cdot100}\right)\)
\(B=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{98\cdot99}-\frac{1}{99\cdot100}\right)\)
\(B=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{99\cdot100}\right)=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{9900}\right)\)
Số hơi bị dữ nên tính nốt nhé
S
30 tháng 7 2018
\(=\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)-\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{99.100.101}\right)\)
\(=\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)-\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{99.100}-\frac{1}{100.101}\right)\)
\(=\left(1-\frac{1}{100}\right)-\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{100.101}\right)\)
\(=\frac{99}{100}-\frac{1}{2}\cdot\frac{5049}{10100}=\frac{99}{100}-\frac{5049}{20200}=\frac{14949}{20200}\)
ND
Nguyễn Đức Trí
VIP
12 tháng 8 2023
a)\(...A=\dfrac{2^{50+1}-1}{2-1}=2^{51}-1\)
b) \(...\Rightarrow B=\dfrac{3^{80+1}-1}{3-1}=\dfrac{3^{81}-1}{2}\)
c) \(...\Rightarrow C+1=1+4+4^2+4^3+...+4^{49}\)
\(\Rightarrow C+1=\dfrac{4^{49+1}-1}{4-1}=\dfrac{4^{50}-1}{3}\)
\(\Rightarrow C=\dfrac{4^{50}-1}{3}-1=\dfrac{4^{50}-4}{3}=\dfrac{4\left(4^{49}-1\right)}{3}\)
Tương tự câu d,e,f bạn tự làm nhé
NM
1
28 tháng 9 2015
A = 1.2. + 2.3 + 3.4 + ... + 99.100
3A = 1.2.3 + 2.3.(4-1) + ... + 99.100.(101-98)
3A = 1.2.3 + 2.3.4 - 2.3.1 + ... + 99.100.101 - 99.100.98
3A = 99.100.101
3A = 999900
A = 333300
BB
30 tháng 9 2018
Bài 1 Số số hạng của dãy là : (50-1):1+1=50(số hạng )
S = (50+1) x 50 : 2 = 1275
ZH
1
18 tháng 9 2015
Câu 1:
Đặt S = 1.2+2.3+3.4+...+30.31
3 S = 1.2.3+2.3.3+3.4.3+...+30.31.3
3 S = 1.2.(3-0) + 2.3.(4-1) + 3.4.(5-2) + ...+ 30.31.(32-29)
3S = 1.2.3 + 2.3.4-2.3 + 3.4.5-2.3.4 + ...+ 30.31.32-29.30.31
3S= 30.31.32
S= 30.31.32/3
\(A=4+4^2+4^3+....+4^{99}+4^{100}\)
\(=4\left(4+1\right)+4^3\left(4+1\right)+...+4^{99}\left(4+1\right)\)
\(=4\cdot5+4^3\cdot5+...+4^{99}\cdot5\)
\(=5\left(4+4^3+...+4^{99}\right)\)
\(S=1\cdot2+2\cdot3+3\cdot4+...+2018\cdot2019\)
\(3S=1\cdot2\cdot3+2\cdot3\cdot3+3\cdot3\cdot4+...+2018\cdot2019\cdot3\)
\(3S=1\cdot2\cdot\left(3-0\right)+2\cdot3\left(4-1\right)+....+2018\cdot2019\left(2020-2017\right)\)
\(3S=1\cdot2\cdot3-0\cdot1\cdot2+2\cdot3\cdot4-1\cdot2\cdot3+....+2018\cdot2019\cdot2020-2017\cdot2018\cdot2019\)
\(3S=2018\cdot2019\cdot2020\)
\(S=\frac{2018\cdot2019\cdot2020}{3}\)
\(1\cdot2\cdot3+2\cdot3\cdot4+...+48\cdot49\cdot50\)
\(4P=1\cdot2\cdot3\cdot4+2\cdot3\cdot4\cdot4+...+48\cdot49\cdot50\cdot4\)
\(4P=1\cdot2\cdot3\left(4-0\right)+2\cdot3\cdot4\left(5-1\right)+....+48\cdot49\cdot50\left(51-47\right)\)
\(4P=1\cdot2\cdot3\cdot4-0\cdot1\cdot2\cdot3+2\cdot3\cdot4\cdot5-1\cdot2\cdot3\cdot4+....+48\cdot49\cdot50\cdot51-47\cdot48\cdot49\cdot50\)
\(P=\frac{48\cdot49\cdot50\cdot51}{4}\)
\(Q=1^2+2^2+3^2+....+113^2\)
\(Q=1\left(2-1\right)+2\left(3-1\right)+....+133\left(134-1\right)\)
\(Q=\left(1\cdot2+2\cdot3+133\cdot134\right)-\left(1+2+3+...+133\right)\)
Áp dụng công thức cho nó nhanh:\(1\cdot2+2\cdot3+...+133\cdot134=\frac{133\cdot134\cdot135}{3}\)
\(1+2+3+...+133=\frac{133\cdot134}{2}\)
Đến đây đưa casio ra mak tính