Đặt \(A=1^2+3^2+...+99^2\)
Công thức tổng quát:
\(1^2+3^2+5^2+...+\left(2n-1\right)^2\)
\(=\dfrac{n.\left(4n^2-1\right)}{3}\)
\(\Rightarrow A=1^2+3^2+5^2+...+99^2\)
Từ \(2n-1=99\)
\(\Rightarrow n=50\)
\(\Rightarrow A=\dfrac{50.\left(4.50^2-1\right)}{3}\)
\(\Rightarrow A=166650\)
Tính giá trị biểu thức A , biết rằng A = M : N
Mà M = \(\dfrac{\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+\dfrac{4}{96}+...+\dfrac{97}{3}+\dfrac{98}{2}+\dfrac{99}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)
N = \(\dfrac{92-\dfrac{1}{9}-\dfrac{2}{10}-\dfrac{3}{11}-...-\dfrac{90}{98}-\dfrac{91}{99}-\dfrac{92}{100}}{\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{495}+\dfrac{1}{500}}\)
Tính tổng
1,A=a1+a2+...+an với ai-ai-1=k
2,B=1/1*2*3+1/2*3*4+...+1/48*49*50
3,C=1+1/3+1/5+...+1/97+1/99
/
1/1*99+1/3*97+...+1/97*3+1/99*1
CMR : 1/3 - 2/3^2 + 3^3 - 4/3^4 + .... + 99/3^99 - 100/3^100 < 3/16
Tính tổng sau
1) B= 1.2+2.3+3.4+......+99.100
2) C= \(1^2+2^2+3^2+...+99^2\)
3) D= \(\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right).....\left(1-\frac{1}{n^2}\right)\)
4) E=\(\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+....+\frac{1}{3^{100}}\)
tính :1*2-1/2!-3*2-1/3!+3*4-1/4!+...+99*100-1/100!
Tính ta được
CMR: \(C=\dfrac{1}{3}+\dfrac{2}{3^2}+\dfrac{3}{3^3}+...+\dfrac{99}{3^{99}}+\dfrac{100}{3^{100}}< \dfrac{3}{4}\)
chứng minh rằng 1/3+1/3^2+1/3^3+...+1/3^99<1/2
Cho biểu thức \(C=\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\)
Chứng minh \(C< \dfrac{3}{16}\)
Tính tổng :
a) 12+22+32+42+...992+1002
b) 12+32+52+72+...972+992
