B = 1 + 5 + 52 + 53 + ....... + 52008 + 52009
S = 1 + 2 + 5 + 14 + ....... + 3n-1 + 1/2 (với n thuộc Z)
A = 1 + 3/2^3 + 4/2^4 + 5/2^5 + ...... + 100/2^100
Q = 1 + 1/2*(1+2) + 1/3*(1+2+3) + 1/4*(1+2+3+4) + ...... + 1/20*(1+2+3+.....+20)
M = -4/1*5 - 4/5*9 - 4/9*13 - ....... - 4/(n+4)*n
Giúp mk với! Mk đang cần gấp lắm !!!!!
\(B=1+5+5^2+5^3+...+5^{2008}+5^{2009}\)
\(\Rightarrow 5B=5+5^2+5^3+5^4+...+5^{2009}+5^{2010}\)
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 \(S=\frac{3^n-1}{4}+\frac{n}{2}\)
\(A=1+\frac{3}{2^3}+\frac{4}{2^4}+\frac{5}{2^5}+...+\frac{100}{2^{100}}\)
\(\Rightarrow 2A=2+\frac{3}{2^2}+\frac{4}{2^3}+\frac{5}{2^4}+...+\frac{100}{2^{99}}\)
Trừ theo vế:
\(2A-A=1+\frac{3}{2^2}+\frac{4-3}{2^3}+\frac{5-4}{2^4}+\frac{6-5}{2^5}+...+\frac{100-99}{2^{99}}-\frac{100}{2^{100}}\)
\(\Leftrightarrow A=1+\frac{3}{4}-\frac{100}{2^{100}}+(\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{99}})\)
Đặt \(T=(\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{99}})\)
\(\Rightarrow 2T=\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{98}}\)
Trừ theo vế: \(2T-T=\frac{1}{2^2}-\frac{1}{2^{99}}\)
\(\Leftrightarrow T=\frac{1}{4}-\frac{1}{2^{99}}\)
Do đó: \(A=1+\frac{3}{4}-\frac{100}{2^{100}}+\frac{1}{4}-\frac{1}{2^{99}}=2-\frac{102}{2^{100}}\)
Áp dụng công thức \(1+2+...+n=\frac{n(n+1)}{2}\) ta có:
\(Q=1+\frac{1}{2}.(1+2)+\frac{1}{3}(1+2+3)+\frac{1}{4}(1+2+3+4)+...+\frac{1}{20}(1+2+3+...+20)\)
\(=1+\frac{1}{2}.\frac{2(2+1)}{2}+\frac{1}{3}.\frac{3(3+1)}{2}+\frac{1}{4}.\frac{4(4+1)}{2}+...+\frac{1}{20}.\frac{20(20+1)}{2}\)
\(=1+\frac{3}{2}+\frac{4}{2}+...+\frac{21}{2}=1+\frac{3+4+5+...+21}{2}\)
\(=1+\frac{1+2+3+4+...+21}{2}-\frac{1+2}{2}\)
\(=1+\frac{21(21+1)}{4}-\frac{3}{2}=115\)
\(M=-\frac{4}{1.5}-\frac{4}{5.9}-...-\frac{4}{(n+4)n}\)
\(\Rightarrow -M=\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{n(n+4)}\)
\(-M=\frac{5-1}{1.5}+\frac{9-5}{5.9}+\frac{13-9}{9.13}+...+\frac{(n+4)-n}{n(n+4)}\)
\(-M=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{n}-\frac{1}{n+4}\)
\(-M=1-\frac{1}{n+4}=\frac{n+3}{n+4}\)
\(\Rightarrow M=\frac{-(n+3)}{n+4}\)
