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C/m nó nhỏ hơn 3/4 hả bạn ?
Có \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}< \frac{1}{4}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=\frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{4}+\frac{1}{2}-\frac{1}{100}< \frac{3}{4}\)
A = 1.2 + 2.3 + 3.4 + ...+ 59.60
3A = 1.2.3 + 2.3.3 + 3.4.3 + ...+ 59.60.3
3A = 1.2.(3-0) + 2.3.(4-1) + 3.4.(5-2) +...+ 59.60.(61-58)
3A = 1.2.3 - 0 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 +...+ 59.60.61 - 58.59.60
3A = 58.59.60 => A = 58.59.60 : 3 = 68 440
B = 12 + 22 + 32 + 592
B = 1.2 + 2.2 + 3.3 + 59.59
B = 2 + 4 + 9 + 3481
B = 3496
vậy A - B = 68 440 - 3 496 = 64 944
( bấm nhé )
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{100.100}\)
\(< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}< 1\)(ĐPCM)
A=1/4(1/1+1/2^2+...+1/50^2)
=>A=1/4+1/4*(1/2^2+...+1/50^2)
=>A<1/4+1/4*(1-1/2+1/2-1/3+...+1/49-1/50)
=>A<1/4+1/4*49/50=99/200<1/2
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2020^2}\)
\(< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\)
\(=1-\frac{1}{2020}< 1\)
\(3^6:3^2+2^3.2^2-3^3.3\)
\(=3^4+2^5-3^4\)
\(=3^4-3^4+2^5\)
\(=0+2^5=2^5\)
\(3^6:3^2+2^3.2^2-3^3.3\\ =3^4+2-3^4\\ =\left(3^4-3^4\right)+2\\ =0+2\\ =2.\)
a)31x32x33x........x3100
=31+2+3+4+...+100
=3(100+1)x(100-1+1):2
=3101x100:2
=35050
Bài b mình không biết làm
Ta có:
\(n^2=n^2-n+n=n\left(n-1\right)+n\)
Áp dụng ta được:
\(A=1^2+2^2+3^2+...+100^2\)
\(=1.2+2.3+...+99.100+1+2+3+...+100\)
Đặt \(B=1.2+2.3+...+99.100\)
\(C=1+2+3+...+100\)
\(3B=1.2.3+2.3.3+...+99.100.3\)
\(=1.2.3+2.3.\left(4-1\right)+...+99.100.\left(101-98\right)\)
\(=1.2.3+2.3.4-1.2.3+...+99.100.101-98.99.100\)
\(=99.100.101\)
\(B=\frac{99.100.101}{3}=100.101.33\)
\(C=1+2+...+100=\frac{100.101}{2}\)
\(A=B+C=100.101.33+\frac{100.101}{2}=\frac{100.101.\left(200+1\right)}{6}\)