Tính M = 1x2+2x3+3x4+....+212x213
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ta có công thức:1x2+2x3+3x4...+Nx(N+1)=(Nx(N+1)x(N+2)):3
suy ra M=(212x(212+1)x(212+2)):3=3221128
\(\Rightarrow3M=1.2.3+2.3.3+...+201.202.3\)
\(=1.2.3+2.3.\left(4-1\right)+...+201.202.\left(203-200\right)\)
\(=1.2.3+2.3.4-1.2.3+...+201.202.203-200.201.202\)
\(=201.202.203\)
\(\Rightarrow M=\frac{201.202.203}{3}\)
\(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+....+\dfrac{1}{24\times25}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{24}-\dfrac{1}{25}\)
\(=1-\dfrac{1}{25}\)
\(=\dfrac{24}{25}\)
A= 1x2+2x3+3x4+...+98x99 A x 3= 1x2 x (3-0) +2x3x (4-1)+3x4 x (5-2)+...+98x99x (100-97) = 1x2x3+2x3x4+......98x99x100- (1x2x0+ 2x3x1+....+ 98x99x97) = 98x99x100
A= 1x2+2x3+3x4+...+98x99
A x 3= 1x2 x (3-0) +2x3x (4-1)+3x4 x (5-2)+...+98x99x (100-97)
= 1x2x3+2x3x4+......98x99x100- (1x2x0+ 2x3x1+....+ 98x99x97)
= 98x99x100.
1x 2 + 2 x 3 + 3 x 4 + ...+ 99 x 100
Ta có:
1 x 2 x 3 = 1 x 2 x 3
2 x 3 x 3 = 2 x 3 x ( 4 - 1) = 2 x 3 x 4 - 1 x 2 x 3
3 x 4 x 3 = 3 x 4 x ( 5 - 2) = 3 x 4 x 5 - 2 x 3 x 4
........................................................= ........................................
99 x 100 x 3 = 99 x 100 x (101 - 98) = 99 x 100 x 101 - 99 x 100 x 98
Cộng vế với vế ta có:
1 x 2 x 3 + 2 x 3 x 3 + 3 x 4 x 3 +...+ 99 x 100 x 3 = 99 x100 x 101
(1 x 2 + 2 x 3 + 3 x 4 +...+ 99 x 100) x 3 = 99 x 100 x 101
1 x 2 + 2 x 3 + 3 x 4 +...+ 99 x 100 = \(\dfrac{99\times100\times101}{3}\)
1 x 2 + 2 x 3 + 3 x 4 + ....+ 99 x 100 = 333300
Đặt A = 1×2 + 2×3 + 3×4 + ... + 19×20
⇒ 3A = 1×2×3 + 2×3×3 + 3×4×3 + ... + 19×20×3
= 1×2×3 + 2×3×(4 - 1) + 3×4×(5 - 2) + ... + 19×20×(21 - 18)
= 1×2×3 - 1×2×3 + 2×3×4 - 2×3×4 + 3×4×5 - ... - 18×19×20 + 19×20×21
= 19×20×21
= 7980
⇒ A = 7980 : 3 = 2660
\(M=\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{99\times100}\)
\(M=\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+...+\left(\frac{1}{99}-\frac{1}{100}\right)\)
\(M=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(M=\frac{1}{1}-\frac{1}{100}\)
\(M=\frac{100}{100}-\frac{1}{100}\)
\(M=\frac{99}{100}\)
\(M=\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{99\times100}\)
\(M=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{100}-\frac{1}{100}\)
\(M=1-\frac{1}{100}\)
\(M=\frac{99}{100}\)
Ta có : M = 1.2 + 2.3 + 3.4 + ..... + 212.213
=> 3M = 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + ...... + 212.213.214
=> 3M = 212.213.214
=> M = 212.213.214/3
=> M = 3221128