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HT
4 tháng 8 2015
D = \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+....+\frac{1}{37.38.39}\)
D = \(\frac{1}{2}.\left(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+...+\frac{39-37}{37.38.39}\right)\)
D = \(\frac{1}{2}.\left(\frac{3}{1.2.3}-\frac{1}{1.2.3}+\frac{4}{2.3.4}-\frac{2}{2.3.4}+...+\frac{39}{37.38.39}-\frac{37}{37.38.39}\right)\)
D = \(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{37.38}-\frac{1}{38.39}\right)\)
D = \(\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{38.39}\right)\)
D = \(\frac{1}{2}.\frac{370}{741}\)
D = \(\frac{185}{741}\)
29 tháng 11 2016
Đặt A = 1.2.3 + 2.3.4 + 3.4.5 + ... + 28.29.30
4A = 1.2.3.(4-0) + 2.3.4.(5-1) + 3.4.5.(6-2) + ... + 28.29.30.(31-27)
4A = 1.2.3.4 - 0.1.2.3. + 2.3.4.5 - 1.2.3.4 + 3.4.5.6 - 2.3.4.5 + ... + 28.29.30.31 - 27.28.29.30
4A = 28.29.30.31 - 0.1.2.3
4A = 28.29.30.31
\(A=\frac{28.29.30.31}{4}=7.29.30.31=188790\)
Theo cách tính trên ta dễ dàng tính được:
1.2.3 + 2.3.4 + 3.4.5 + ... + (n - 1).n.(n + 1) = \(\frac{\left(n-1\right).n.\left(n+1\right).\left(n+2\right)}{4}\)
H9
HT.Phong (9A5)
CTVHS
3 tháng 8 2023
\(B=1\cdot2\cdot3+2\cdot3\cdot4+3\cdot4\cdot5+...+98\cdot99\cdot100\)
\(\Rightarrow4B=4\cdot\left(1\cdot2\cdot3+2\cdot3\cdot4+...+98\cdot99\cdot100\right)\)
\(\Rightarrow4B=1\cdot2\cdot3\cdot\left(4-0\right)+2\cdot3\cdot4\cdot\left(5-1\right)+3\cdot4\cdot5\cdot\left(6-2\right)+...+98\cdot99\cdot100\cdot\left(101-97\right)\)
\(\Rightarrow4B=1\cdot2\cdot3\cdot4+2\cdot3\cdot4\cdot5-1\cdot2\cdot3\cdot4-....+98\cdot99\cdot100\cdot101-97\cdot98\cdot99\cdot100\)
\(\Rightarrow4B=98\cdot99\cdot100\cdot101\)
\(\Rightarrow B=\dfrac{98\cdot99\cdot100\cdot101}{4}\)
\(\Rightarrow B=25\cdot98\cdot99\cdot101\)
3 tháng 8 2023
B=1x2x3+2x3x4+...+98x99x100
=>4B=1x2x3x(4-0)+2x3x4x(5-1)+...+98x99x100x(101-97)
4B=1x2x3x4+2x3x4x5-1x2x3x4+...+98x99x100x101-97x98x99x100
4B=98x99x100x101
=>B=\(\dfrac{98\cdot99\cdot100\cdot101}{4}\)=24497550.
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
13 tháng 1 2023
A= 1.2.3 + 2.3.4 + 3.4.5 +.....+ 98.99.100
4A = 98.99.100.4 + .....+ 3.4.5.4 + 2.3.4.4 + 1.2.3.4
4A = 98.99.100.(101-97) +... + 2.3.4.(5-1) + 1.2.3.4
4A = 98.99.100.101 - 97.98.99.100+......+2.3.4.5 - 1.2.3.4 + 1.2.3.4
4A = 98.99.100.101
A = 98.99.100.101 : 4
A = 24497550
13 tháng 11 2023
\(B=1\cdot2\cdot3+2\cdot3\cdot4+...+\left(n-1\right)\cdot n\cdot\left(n+1\right)\)
=>\(4B=1\cdot2\cdot3\cdot4+2\cdot3\cdot4\cdot4+...+\left(n-1\right)\cdot n\left(n+1\right)\cdot4\)
=>\(4B=1\cdot2\cdot3\cdot4+2\cdot3\cdot4\left(5-1\right)+...+\left(n-1\right)\cdot n\left(n+1\right)\left[\left(n+2\right)-\left(n-2\right)\right]\)
=>\(4B=1\cdot2\cdot3\cdot4-1\cdot2\cdot3\cdot4+...+\left(n-2\right)\left(n-1\right)\cdot n\cdot\left(n+1\right)-\left(n-2\right)\cdot\left(n-1\right)\cdot n\cdot\left(n+1\right)+\left(n-1\right)\cdot n\left(n+1\right)\left(n+2\right)\)
=>\(4B=\left(n-1\right)\cdot n\cdot\left(n+1\right)\left(n+2\right)\)
=>\(B=\dfrac{\left(n-1\right)\cdot n\left(n+1\right)\left(n+2\right)}{4}\)
\(C=1\cdot4+2\cdot5+3\cdot6+...+n\left(n+3\right)\)
\(=1\cdot\left(1+3\right)+2\left(2+3\right)+...+n\left(n+3\right)\)
\(=\left(1^2+2^2+...+n^2\right)+3\left(1+2+...+n\right)\)
\(=\dfrac{n\left(n+1\right)\left(2n+1\right)}{6}+3\cdot\dfrac{n\left(n+1\right)}{2}\)
\(=\dfrac{n\left(n+1\right)\left(2n+1\right)}{6}+\dfrac{3n\left(n+1\right)}{2}\)
\(=\dfrac{n\left(n+1\right)}{2}\cdot\left(\dfrac{2n+1}{3}+3\right)\)
\(=\dfrac{n\left(n+1\right)}{2}\cdot\dfrac{2n+1+9}{3}\)
\(=\dfrac{n\left(n+1\right)\left(n+5\right)}{3}\)
\(D=1^2+2^2+...+n^2\)
\(=1+\left(1+1\right)\cdot2+\left(1+2\right)\cdot3+...+\left(1+n-1\right)\cdot n\)
\(=1+2+3+...+n+\left(1\cdot2+2\cdot3+...+\left(n-1\right)\cdot n\right)\)
Đặt \(A=1+2+3+...+n;E=1\cdot2+2\cdot3+...+\left(n-1\right)\cdot n\)
\(E=1\cdot2+2\cdot3+...+\left(n-1\right)\cdot n\)
=>\(3E=1\cdot2\cdot3+2\cdot3\cdot3+...+\left(n-1\right)\cdot n\cdot3\)
=>\(3E=1\cdot2\cdot3+2\cdot3\cdot\left(4-1\right)+...+\left(n-1\right)\cdot n\left[\left(n+1\right)-\left(n-2\right)\right]\)
=>\(3E=1\cdot2\cdot3-1\cdot2\cdot3+2\cdot3\cdot4+...+\left(n-1\right)\cdot n\left(n-2\right)-\left(n-1\right)\cdot n\left(n-2\right)+\left(n-1\right)\cdot n\cdot\left(n+1\right)\)
=>\(3E=\left(n-1\right)\cdot n\left(n+1\right)=n^3-n\)
=>\(E=\dfrac{n^3-n}{3}\)
\(A=1+2+3+...+n\)
Số số hạng là n-1+1=n(số)
Tổng của dãy số là: \(A=\dfrac{n\left(n+1\right)}{2}\)
=>\(D=\dfrac{n^3-n}{3}+\dfrac{n\left(n+1\right)}{2}\)
\(=\dfrac{2n^3-2n+3n^2+3n}{6}\)
=>\(D=\dfrac{2n^3+3n^2+n}{6}\)
6 tháng 4 2021
Ta có: \(S=1\cdot2\cdot3+2\cdot3\cdot4+3\cdot4\cdot5+...+97\cdot98\cdot99\)
\(\Leftrightarrow4\cdot S=1\cdot2\cdot3\cdot\left(4-0\right)+2\cdot3\cdot4\cdot\left(5-1\right)+3\cdot4\cdot5\cdot\left(6-2\right)+...+97\cdot98\cdot99\cdot\left(101-97\right)\)
\(\Leftrightarrow4\cdot S=98\cdot99\cdot100\cdot101\)
\(\Leftrightarrow S=\text{24497550}\)
4 tháng 8 2015
4A=4.[1.2.3 + 2.3.4 + 3.4.5 + 4.5.6 + . . . . . . + 48.49.50]
4A=1.2.3.4 +2.3.4.4 +3.4.5.4 +4.5.6.4 +.........+48.49.50.4
4A=1.2.3.4 +2.3.4.(5-1) + 3.4.5.(6-2) +4.5.6(7-3)+....+ 48.49.50(51-47)
4A=1.2.3.4 +2.3.4.5 --1.2.3.4 + 3.4.5.6--2.3.4.5 + 4.5.6.7-3.4.5.6+....+ 48.49.50.51--47.48.49.50
4A =48.49.50.51
A=(48.49.50.51)/4
Câu hỏi của hồ thị hằng - Toán lớp 6 - Học toán với OnlineMath
\(4D=1.2.3.\left(4-0\right)+2.3.4.\left(5-1\right)+.....+98.99.100\left(101-97\right)\)
\(\Rightarrow4D=1.2.3.4+2.3.4.5-1.2.3.4+.....+98.99.100.101-97.98.99.100\)
\(\Rightarrow4D=98.99.100.101\)
\(\Rightarrow D=\frac{98.99.100.101}{4}\)
\(\Rightarrow D=24497550\)