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a) 3A=1.2.3 + 2.3.3 + 3.4.3 +... + n.(n+1).3
=1.2.(3-0) + 2.3.(4-1) + ... + n.(n+1).[(n+2)-(n-1)]
=[1.2.3+ 2.3.4 + ...+ (n-1).n.(n+1)+ n.(n+1)(n+2)] - [0.1.2+ 1.2.3 +...+(n-1).n.(n+1)]
=n.(n+1).(n+2)
=>S=[n.(n+1).(n+2)] /3
b)
Nhân 4 vào hai vế ta được:
4A = 4.[1.2.3 + 2.3.4 + 3.4.5 + … + (n – 1).n.(n + 1)]
4A = 1.2.3.4 + 2.3.4.4 + 3.4.5.4 + … + (n – 1).n.(n + 1).4
4A = 1.2.3.4 + 2.3.4.(5 – 1) + 3.4.5.(6 – 2) + … + (n – 1).n.(n + 1).[(n + 2) – (n – 2)]
4A = 1.2.3.4 + 2.3.4.5 – 1.2.3.4 + 3.4.5.6 – 2.3.4.5 + … + (n – 1).n(n + 1).(n + 2) – (n – 2).(n – 1).n.(n + 1)
4A = (n – 1).n(n + 1).(n + 2)
A = (n – 1).n(n + 1).(n + 2) : 4.
3A=1.2.3 + 2.3.3 + 3.4.3 +... + n.(n+1).3
=1.2.(3-0) + 2.3.(4-1) + ... + n.(n+1).[(n+2)-(n-1)]
=[1.2.3+ 2.3.4 + ...+ (n-1).n.(n+1)+ n.(n+1)(n+2)] - [0.1.2+ 1.2.3 +...+(n-1).n.(n+1)]
=n.(n+1).(n+2)
=>S=[n.(n+1).(n+2)] /3
Ta có ; K = \(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.....+\frac{1}{45}\)
\(=1+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{90}\)
\(=1+\left(\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+.....+\frac{2}{9.10}\right)\)
\(=1+2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.....+\frac{1}{9.10}\right)\)
\(=1+2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{9}-\frac{1}{10}\right)\)
\(=1+2\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(=1+1-\frac{1}{5}\)(nhân phá ngoặc)
\(=2-\frac{1}{5}\)< 2
Vậy K = \(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.....+\frac{1}{45}\)< 2
2, tìm x thuộc Z biết :
a, x^2 -(-3 )^2 =16
x^2-9 = 16
x^2 = 25
=> x = 5
b, x^2 + (-4) ^2 =0
x^2 + 16 = 0
x^2 = -16
=> x= -4
\(1)x+\frac{5}{6}\times2\frac{2}{5}-1\frac{1}{4}=35\%\)
\(x+\frac{5}{6}\times\frac{12}{5}-\frac{5}{4}=\frac{7}{12}\)
\(x+\frac{5}{6}\times\frac{12}{5}=\frac{7}{12}+\frac{5}{4}\)
\(x+\frac{5}{6}.\frac{12}{5}=\frac{8}{5}\)
\(x+\frac{5}{6}=\frac{8}{5}:\frac{12}{5}\)
\(x+\frac{5}{6}=\frac{2}{3}\)
\(x=\frac{2}{3}-\frac{5}{6}\)
\(x=-\frac{1}{6}\)
HỌC TỐT !
\(2\)) \(\left|x-\frac{1}{2}\right|-\frac{3}{4}=0\)
\(\left|x-\frac{1}{2}\right|\) \(=0+\frac{3}{4}\)
\(\left|x-\frac{1}{2}\right|\) \(=\frac{3}{4}\)
\(x-\frac{1}{2}\) \(=\frac{3}{4}\)hoặc \(-\frac{3}{4}\)
Ta xét 2 trường hợp :
Trường hợp 1 : \(x-\frac{1}{2}=\frac{3}{4}\)
\(x\) \(=\frac{3}{4}+\frac{1}{2}\)
\(x\) \(=\frac{5}{4}\)
Trường hợp 2 : \(x-\frac{1}{2}=-\frac{3}{4}\)
\(x\) \(=-\frac{3}{4}+\frac{1}{2}\)
\(x\) \(=-\frac{1}{4}\)
Vậy \(x\in\text{{}\frac{5}{4};-\frac{1}{4}\)}
Ta có : E = 22 + 24 + 26 + ...... + 2100
=> 22E = 24 + 26 + 28 + ........ + 2102
=> 4E = 2102 - 22
=> E = \(\frac{2^{102}-4}{4}\)
A = 4 + (22 + 23 + 24 + ... + 220)
A - 4 = 22 + 23 + 24 + ... + 220
2(A - 4) = 23 + 24 + 25 + ... + 221
A - 4 = 2(A - 4) - (A - 4) = (23 + 24 + 25 + ... + 221) - (22 + 23 + 24 + ... + 220)
A - 4 = (23 - 23) + (24 - 24) + ... + (220 - 220) + (221 - 22)
A - 4 = 221 - 4
A = 221 - 4 + 4
A = 221
A=221.Duyệt nha