Cho A = 1-$\frac{1}{2}$ + $\frac{1}{3}$ – $\frac{1}{4}$ + … – $\frac{1}{2016}$ + $\frac{1}{2017}$ – $\frac{1}{2018}$.
B = $\frac{1}{1010}$ + $\frac{1}{1011}$ +…+ $\frac{1}{2016}$ + $\frac{1}{2017}$ + $\frac{1}{2018}$.
Tính : ($A^{2017}$ – $B^{2017}$) $^{2018}$
Cho A = 1-$\frac{1}{2}$ + $\frac{1}{3}$ – $\frac{1}{4}$ + … – $\frac{1}{2016}$ + $\frac{1}{2017}$ – $\frac{1}{2018}$. B = $\frac{1}{1010}$ + $\fr
By Ximena
Đáp án: 0
Giải thích các bước giải:
A = 1 – $\frac{1}{2}$ + $\frac{1}{3}$ – $\frac{1}{4}$ +…- $\frac{1}{2016}$ + $\frac{1}{2017}$ – $\frac{1}{2018}$
= (1 + $\frac{1}{2}$ + $\frac{1}{3}$ + $\frac{1}{4}$ +…- $\frac{1}{2016}$ + $\frac{1}{2017}$ – $\frac{1}{2018}$) – 2( $\frac{1}{2}$ +$\frac{1}{3}$ + $\frac{1}{4}$ +…- $\frac{1}{2016}$ + $\frac{1}{2017}$ – $\frac{1}{2018}$)
= (1 + $\frac{1}{2}$ + $\frac{1}{3}$ +…- $\frac{1}{2016}$ + $\frac{1}{2017}$ – $\frac{1}{2018}$) – (1 + $\frac{1}{2}$ +…+$\frac{1}{1009}$)
= $\frac{1}{1010}$ + $\frac{1}{1011}$ +…+ $\frac{1}{2016}$ + $\frac{1}{2017}$ + $\frac{1}{2018}$
= B
⇒ $A^{2017}$ = $B^{2017}$
($A^{2017}$ – $B^{2017}$)^2018 = $0^{2018}$ = 0