\(-\frac{1}{2}-\frac{1}{3}-\frac{1}{4}-...-\frac{1}{20}+\frac{3}{2}+\frac{4}{3}+\frac{5}{4}+...+\frac{21}{20}\)

S=-1/2-1/3-1/4-...-1/20+3/2+4/3+5/4+...+21/20

=>S=(3/2-1/2)+(4/3-1/3)+(5/4-1/4)+...+(21/20-1/20)

=>S=1+1+1...+1

Ta thấy S có 20 số hạng

=>S=20

\(B=1+5+5^2+5^3+...+5^{2008}+5^{2009}\)

\(\Rightarrow 5B=5+5^2+5^3+5^4+...+5^{2009}+5^{2010}\)

Trừ theo vế:

\(5B-B=(5+5^2+5^3+5^4+...+5^{2009}+5^{2010})-(1+5+5^2+...+5^{2009})\)

\(4B=5^{2010}-1\)

\(B=\frac{5^{2010}-1}{4}\)

\(S=\frac{3^0+1}{2}+\frac{3^1+1}{2}+\frac{3^2+1}{2}+..+\frac{3^{n-1}+1}{2}\)

\(=\frac{3^0+3^1+3^2+...+3^{n-1}}{2}+\frac{\underbrace{1+1+...+1}_{n}}{2}\)

\(=\frac{3^0+3^1+3^2+..+3^{n-1}}{2}+\frac{n}{2}\)

Đặt \(X=3^0+3^1+3^2+..+3^{n-1}\)

\(\Rightarrow 3X=3^1+3^2+3^3+...+3^{n}\)

Trừ theo vế:

\(3X-X=3^n-3^0=3^n-1\)

\(\Rightarrow X=\frac{3^n-1}{2}\). Do đó \(S=\frac{3^n-1}{4}+\frac{n}{2}\)

19 nha pạn tick nhé bạn hiền

\(\frac{-1}{2}-\frac{1}{3}-\frac{1}{4}-.........-\frac{1}{20}+\frac{3}{2}+\frac{4}{3}+\frac{5}{4}+...........+\frac{21}{20}\)

=\(\left(\frac{-1}{2}+\frac{3}{2}\right)+\left(\frac{-1}{3}+\frac{4}{3}\right)+\left(\frac{-1}{4}+\frac{5}{4}\right)+..................+\left(\frac{-1}{20}+\frac{21}{20}\right)\)

=\(1+1+1+.........+1\)(19 số 1)

=19

Ta có:

\(B=1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+...+\dfrac{1}{20}\left(1+2+...+20\right)\)

\(=1+\dfrac{1}{2}.\dfrac{2\left(2+1\right)}{2}+\dfrac{1}{3}.\dfrac{3\left(3+1\right)}{2}+...+\dfrac{1}{20}.\dfrac{20\left(20+1\right)}{2}\)

\(=\dfrac{2}{2}+\dfrac{2+1}{2}+\dfrac{3+1}{2}+...+\dfrac{20+1}{2}\)

\(=\dfrac{2}{2}+\dfrac{3}{2}+\dfrac{4}{2}+...+\dfrac{20}{2}\)

\(=\dfrac{2+3+4+...+20}{2}=\dfrac{\dfrac{20\left(20+1\right)}{2}-1}{2}\)

\(=\dfrac{209}{2}\)

Vậy \(B=\dfrac{209}{2}\)

Sao 20+1/2 ở trên xuống dưới thành 20/2 vậy.