TÍNH
A)$\frac{1}{3^2}$-$\frac{1}{3^3}$ +…+$\frac{1}{3mũ50}$ -$\frac{1}{3mũ51}$
B) $\frac{1}{1.2.3}$ +$\frac{1}{2.3.4}$ +…+$\frac{1}{a.(a+1).(a+2)}$
TÍNH
A)$\frac{1}{3^2}$-$\frac{1}{3^3}$ +…+$\frac{1}{3mũ50}$ -$\frac{1}{3mũ51}$
B) $\frac{1}{1.2.3}$ +$\frac{1}{2.3.4}$ +…+$\frac{1}{a.(a+1).(a+2)}$
Đáp án:
`dowmnarrow`
Giải thích các bước giải:
`A=1/3^2-1/3^3+……+1/3^50-1/3^51`
`3A=1/3-1/3^2+…..+1/3^49-1/3^50`
`=>3A+A=4A=1/3-1/3^51`
`=>A=(1/3-1/3^51)/4`
`B=1/(1.2.3)+1/(2.3.4)+……..+1/(a(a+1)(a+2))`
`=>2A=2/(1.2.3)+2/(2.3.4)+……..+2/(a(a+1)(a+2))`
`=>2A=1/(1.2)-1/(2.3)+1/(2.3)-1/(3.4)+…….+1/(a.(a+1))-1/(a(a+2))`
`=>2A=1/2-1/(a^2+2a)`
`=>A=(1/2-1/(a^2+2a))/2`
a)
A=$\frac{1}{3^2}$ – $\frac{1}{3^3}$ + … + $\frac{1}{3^{50}}$ – $\frac{1}{3^{51}}$
=> 3A = $\frac{1}{3}$ – $\frac{1}{3^2}$ + $\frac{1}{3^3}$ – …+ $\frac{1}{3^{49}}$ – $\frac{1}{3^{50}}$
=> 3A + A= ($\frac{1}{3^2}$ – $\frac{1}{3^3}$ + … + $\frac{1}{3^{50}}$ – $\frac{1}{3^{51}}$) + ($\frac{1}{3}$ – $\frac{1}{3^2}$ + $\frac{1}{3^3}$ – …+ $\frac{1}{3^{49}}$ – $\frac{1}{3^{50}}$ )
Rút gọn đc:
4A= $\frac{1}{3}$ – $\frac{1}{3^{51}}$
4A= $\frac{3^{51}}{3^{52}}$ – $\frac{3}{3^{52}}$ (Quy đồng mẫu)
4A= $\frac{3^{51} – 3}{3^{52}}$
=>A= $\frac{3^{51} – 3}{3^{52}}$ . $\frac{1}{4}$
A= $\frac{3^{51} – 3}{3^{52} .4}$
b) B=$\frac{1}{1.2.3}$ + $\frac{1}{2.3.4}$ +…+ $\frac{1}{a.(a+1).(a+2)}$
=>2B= $\frac{2}{1.2.3}$ + $\frac{2}{2.3.4}$ +…+ $\frac{2}{a.(a+1).(a+2)}$
=>2B=($\frac{1}{1.2}$ – $\frac{1}{2.3}$) + ($\frac{1}{2.3}$ – $\frac{1}{3.4}$) + … + ($\frac{1}{a.(a+1)}$ – $\frac{1}{(a+1).(a+2)}$)
Rút gọn đc:
2B=$\frac{1}{2}$ – $\frac{1}{(a+1).(a+2)}$
2B=$\frac{(a+1).(a+2) – 2}{2.(a+1).(a+2)}$
=>B=$\frac{(a+1).(a+2) – 2}{4.(a+1).(a+2)}$