P = 1/49+2/48+3/47+...+48/2+49/1
Cộng 1 váo mỗi p/s trong 48 p/s đầu , trừ p/s cuối đi 48 ta được
P=(1/49+1)+(2/48+1)+...+(48/2+1)+1
P= 50/49+50/48+....+50/2+50/50
Đưa ps cuối lên đầu
P=50/50+50/49+50/48+...+50/2
=50.(1/50+1/49+1/48+...+1/4+1/3+1/2)
=50S
=> S/P=1/50
Cho S = \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{48}+\dfrac{1}{49}+\dfrac{1}{50}\)và P = \(\dfrac{1}{49}+\dfrac{2}{48}+\dfrac{3}{47}+...+\dfrac{48}{2}+\dfrac{49}{1}\). Tính \(\dfrac{S}{P}\)
S=\(\dfrac{1}{2}\)+\(\dfrac{1}{3}\)+\(\dfrac{1}{4}\)+...+\(\dfrac{1}{49}\)+\(\dfrac{1}{50}\)
P=\(\dfrac{1}{49}\)+\(\dfrac{2}{48}\)+\(\dfrac{3}{47}\)+...+\(\dfrac{48}{2}\)+\(\dfrac{49}{1}\)
Tính \(\dfrac{S}{P}\)
cho A= \(\dfrac{1}{2}\) + \(\dfrac{1}{3}\) + .............+ \(\dfrac{1}{50}\)
B= \(\dfrac{49}{1}\) + \(\dfrac{48}{2}\) + .....+ \(\dfrac{1}{49}\) = 50 * B
CMR A và B ko là số tự nhiên
1 CM
a, \(\left(\dfrac{1}{1}+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{2n-1}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2n}\right)=\dfrac{1}{n+1}+\dfrac{1}{n+2}+...+\dfrac{1}{2n}\)( n∈Z)
b, \(\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{50}=\dfrac{99}{50}-\dfrac{97}{49}+...+\dfrac{7}{4}-\dfrac{5}{3}+\dfrac{3}{2}\)
tính S=\(\dfrac{1}{30}+\dfrac{2}{48}+\dfrac{3}{88}+\dfrac{4}{165}+\dfrac{5}{300}\)
Tính nhanh:
a) A= 1 + \(\dfrac{1}{5}+\dfrac{1}{25}+\dfrac{1}{125}+\dfrac{1}{625}+...+\dfrac{1}{78125}\)
b) B= \(\dfrac{1}{3}+\dfrac{1}{12}+\dfrac{1}{48}+\dfrac{1}{192}+\dfrac{1}{768}+...+\dfrac{1}{36864}\)
c) M= \(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+...+\dfrac{1}{9900}\)
d) P= \(\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+\dfrac{1}{1+2+3+4}+...+\dfrac{1}{1+2+3+4+...+2018}\)
1,4 . \(\dfrac{15}{49}\) - (\(\dfrac{4}{5}\) + \(\dfrac{-2}{3}\)) : \(2\dfrac{1}{5}\)
Cho S=\(\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+\dfrac{4}{4^4}+...+\dfrac{2023}{4^{2023}}\). Chứng minh S < \(\dfrac{1}{2}\)
a, Cho b là số tự nhiên, b>1. Chứng minh rằng: \(\dfrac{1}{b}-\dfrac{1}{b+1}< \dfrac{1}{b^2}< \dfrac{1}{b-1}-\dfrac{1}{b}\)
b, Áp dụng phần a: Cho S\(=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}\). Chứng minh rằng: \(\dfrac{2}{5}< S< \dfrac{8}{9}\)