chung minh rang \(\frac{1\cdot3\cdot5\cdot\cdot\cdot39}{21\cdot22\cdot23\cdot\cdot\cdot40}=\frac{1}{2^{20}}\)

\(\frac{1\cdot3\cdot5\cdot...\cdot39}{21\cdot22\cdot23\cdot...\cdot40}\)=\(\frac{1}{2^{20}}\) 

Chứng minh:

\(\frac{1\cdot3\cdot5\cdot...\cdot39}{21\cdot22\cdot23\cdot...40}\)=\(\frac{1}{2^{20}}\)

Chứng minh rằng:

a)\(\frac{1\cdot3\cdot5\cdot\cdot\cdot39}{21\cdot22\cdot23\cdot\cdot\cdot40}=\frac{1}{2^{20}}\)

b)\(\frac{1\cdot3\cdot5\cdot\cdot\cdot\left(2n-1\right)}{\left(n+1\right)\left(n+2\right)\left(n+3\right)\cdot\cdot\cdot2n}=\frac{1}{2^n}\)Với \(n\inℕ^∗\)

\(\frac{1414+1515+1616+1717+1818+1919}{20\cdot20+21\cdot21+22\cdot22+23\cdot23+24\cdot24+25\cdot25}\)Tính nhanh

Giải đầyđủ nha mình cần gấp lắm hôm nay phải có. Mơn nhìu <3

tính tổng sau:

\(A=1\cdot2\cdot3\cdot.....\cdot2021-1\cdot2\cdot3\cdot.....\cdot2021-1\cdot2\cdot3\cdot.....\cdot2019\cdot2020^2\)

chứng tỏ \(\frac{1}{1\cdot2}+\frac{1}{1\cdot2\cdot3}+\frac{1}{1\cdot2\cdot3\cdot4}+...+\frac{1}{1\cdot2\cdot3\cdot...\cdot100}< 1\)

N=\(\frac{5\cdot\left(2^2\cdot3^2\right)\cdot\left(2^2\right)^6-2\cdot\left(2^2\cdot3\right)^{14}\cdot3^6}{5\cdot2^{28}\cdot3^{19}-7\cdot2^{29}\cdot2^{18}}\)

Tinh N

\(A=\frac{1\cdot2}{2\cdot2}\cdot\frac{2\cdot3}{3\cdot3}\cdot\frac{3\cdot4}{4\cdot4}\cdot\frac{4\cdot5}{5\cdot5}\cdot.................\cdot\frac{2012\cdot2013}{2013\cdot2013}\)với

\(B=\frac{2012\cdot2013-2012\cdot2012}{2012\cdot2011+2012\cdot2}\)