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A = 1.2 + 2.3 + 3.4 + ... + 98.99
A = 1.(1 + 1) + 2.(2 + 1) + 3.(3 + 1) + ... + 98.(98 + 1)
A = 12 + 1 + 22 + 2 + 32 + 3 + ... + 982.98
A = (12 + 22 + 32 + ... + 982) + (1 + 2 + 3 + ... + 98)
A = (12 + 22 + 32 + ... + 982) + 4851 (1)
B = 12 + 22 + 32 + ... + 982 (2)
(1)(2) => A - B = 4851 ⋮ 4851
ta có: B = 12 + 22 + 32 +...+982 = 1.1 +2.2+3.3+...+98.98
=> A-B = (1.2+2.3+3.4+4.5+...+98.99) - (1.1+2.2+3.3+...+98.98)
A-B = (1.2-1.1) + (2.3-2.2) + (3.4-3.3) + (4.5-4.4) + ...+ (98.99-98.98)
A-B = 1.(2-1) + 2.(3-2) +3.(4-3) + 4.(5-4) + ...+ 98.(99-98)
A-B = 1 +2+3+4+...+98
A-B = (1+98).98:2
A -B = 4851 chia hết cho 4851
C1 : B=\(\frac{1^2}{1.2}.\frac{2^2}{2.3}......\frac{98^2}{98.99}\)=\(\frac{1.1}{1.2}.\frac{2.2}{2.3}......\frac{98.98}{98.99}\)=\(\left(\frac{1.2......98}{1.2.....98}\right).\left(\frac{1.2......98}{2.3......99}\right)\)
\(1.\frac{1}{99}=\frac{1}{99}\)
C2:Đầu tiên cũng tách ra:\(1^2\)=1.1;\(2^2\)=2.2;...;\(98^2\)=98.98
Xong rút gọn ở tử và mẫu được:\(\frac{1}{2}.\frac{2}{3}.......\frac{98}{99}=\frac{1.2.....98}{2.3.....99}=\frac{1}{99}\)
Bạn thấy cách nào rễ hiểu hơn thì ghi nhé
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\(A=4+4^2+4^3+....+4^{99}+4^{100}\)
\(=4\left(4+1\right)+4^3\left(4+1\right)+...+4^{99}\left(4+1\right)\)
\(=4\cdot5+4^3\cdot5+...+4^{99}\cdot5\)
\(=5\left(4+4^3+...+4^{99}\right)\)
\(S=1\cdot2+2\cdot3+3\cdot4+...+2018\cdot2019\)
\(3S=1\cdot2\cdot3+2\cdot3\cdot3+3\cdot3\cdot4+...+2018\cdot2019\cdot3\)
\(3S=1\cdot2\cdot\left(3-0\right)+2\cdot3\left(4-1\right)+....+2018\cdot2019\left(2020-2017\right)\)
\(3S=1\cdot2\cdot3-0\cdot1\cdot2+2\cdot3\cdot4-1\cdot2\cdot3+....+2018\cdot2019\cdot2020-2017\cdot2018\cdot2019\)
\(3S=2018\cdot2019\cdot2020\)
\(S=\frac{2018\cdot2019\cdot2020}{3}\)
\(1\cdot2\cdot3+2\cdot3\cdot4+...+48\cdot49\cdot50\)
\(4P=1\cdot2\cdot3\cdot4+2\cdot3\cdot4\cdot4+...+48\cdot49\cdot50\cdot4\)
\(4P=1\cdot2\cdot3\left(4-0\right)+2\cdot3\cdot4\left(5-1\right)+....+48\cdot49\cdot50\left(51-47\right)\)
\(4P=1\cdot2\cdot3\cdot4-0\cdot1\cdot2\cdot3+2\cdot3\cdot4\cdot5-1\cdot2\cdot3\cdot4+....+48\cdot49\cdot50\cdot51-47\cdot48\cdot49\cdot50\)
\(P=\frac{48\cdot49\cdot50\cdot51}{4}\)
\(Q=1^2+2^2+3^2+....+113^2\)
\(Q=1\left(2-1\right)+2\left(3-1\right)+....+133\left(134-1\right)\)
\(Q=\left(1\cdot2+2\cdot3+133\cdot134\right)-\left(1+2+3+...+133\right)\)
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
A=1.22+2.32+..............+(n-1).n2
A=1.2.2+2.3.3+.......+(n-1).n.n
A=1.2.(3-1)+2.3.(4-1)+.........+(n-1).n.(n+1-1)
A=1.2.3-1.2+2.3.4-2.3+..........+(n-1).n.(n+1)-(n-1).n
A=[1.2.3+2.3.4+.........+(n-1).n.(n+1)]-[1.2+2.3+............+(n-1).n)
Bạn tự làm tiếp nhá
a)A=\(\frac{\left(8+100\right).\left[\left(100-8\right):4+1\right]}{2}=\frac{108.242}{2}=13068\)
b) \(5B=5^2+5^3+...+5^{101}\)
\(5B-B=5^{101}-5\)
\(B=\frac{5^{101}-5}{4}\)
a/ \(3A=1.2.3+2.3.3+3.4.3+4.5.3+...+29.30.3.\)
\(3A=1.2.3+2.3\left(4-1\right)+3.4.\left(5-2\right)+4.5\left(6-3\right)+...+29.30\left(31-28\right)\)
\(3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+4.5.6-3.4.5+...+29.30.31-28.29.30\)
\(3A=29.30.31\Rightarrow A=\frac{29.30.31}{3}=10.29.31=8990\)
c/ \(C=1+2\left(1+1\right)+3\left(2+1\right)+4\left(3+1\right)+...+30\left(29+1\right)\)
\(C=1+2+1.2+2.3+3+3.4+4+...+29.30+30\)
\(C=\left(1+2+3+4+...+30\right)+\left(1.2+2.3+3.4+...+29.30\right)\)
Dấu ngoặc thứ nhất là tính tổng 1 cấp số cộng, dấu ngoặc thứ 2 chính là câu a
b/ Câu b dãy viết ngắn quá chưa tìm ra quy luật
a) A = 1.2 + 2.3 + ... + 29.30
=> 3A = 1.2.3 + 2.3.(4-1) + ... + 29.30.(31-28)
= 1.2.3 + 2.3.4 - 1.2.3 + ... + 29.30.31 - 28.29.30
= 29.30.31
=> A = \(\frac{29.30.31}{3}=8990\)
chi tiết hơn được ko bạn?