Question

[Lan Anh Nguyễn]

Mn oi giup minh bai nay voi, toi minh di hoc roi

a, Tinh A = 1/3 + 1/3^2 + 1/3^3 + ..... +1/3^n

b, Tị so tu nhien co 2 chu so biet so do nhan voi 75 duoc tich la mot so chinh phuong

c, CMR: 1/3 + 2/3^2 +3/3^3 + ... + 101/3^101

Answer 1

[soyeon_Tiểubàng giải]

a) \(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^n}\)

\(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{n-1}}\)

\(3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{n-1}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^n}\right)\)

\(2A=1-\frac{1}{3^n}\)

\(A=\frac{1-\frac{1}{3^n}}{2}\)

b) Gọi số cần tìm là ab (a khác 0; a,b là các chữ số)

Ta có: ab.75 = x² \(\left(x\ne0\right)\)

=> ab.3.5² = x²

Để ab.75 là 1 số chính phương thì ab = 3.k² \(\left(k\ne0\right)\)

Lại có: 9 < ab < 100 => 9 < 3.k² < 100

=> 3 < k² < 34

Mà k² là số chính phương nên \(k^2\in\left\{4;9;16;25\right\}\)

\(\Rightarrow ab\in\left\{12;27;48;75\right\}\)

Vậy số cần tim là 12; 27; 48; 75

c) Đặt \(B=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{101}{3^{101}}\)

\(3B=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{101}{3^{100}}\)

\(3B-B=\left(1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{101}{3^{100}}\right)-\left(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{101}{3^{101}}\right)\)

\(2B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{100}}-\frac{101}{3^{101}}\)

\(6B=3+1+\frac{1}{3}+...+\frac{1}{3^{99}}-\frac{101}{3^{100}}\)

\(6B-2B=\left(3+1+\frac{1}{3}+...+\frac{1}{3^{99}}-\frac{101}{3^{100}}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{100}}-\frac{101}{3^{101}}\right)\)

\(4B=3-\frac{101}{3^{100}}-\frac{1}{3^{100}}+\frac{101}{3^{101}}\)

\(4B=3-\frac{303}{3^{101}}-\frac{3}{3^{101}}+\frac{101}{3^{101}}\)

\(4B=3-\frac{205}{3^{101}}< 3\)

\(\Rightarrow B< \frac{3}{4}\)

Answer 2

[Lan Anh Nguyễn]

Minh bo sung cau c la tong do be hon 3/4