tính tổng sau S= 2^2+3^2+4^2+…..+2019^2 làm theo cách của lướp 6 nhé 01/08/2021 Bởi Harper tính tổng sau S= 2^2+3^2+4^2+…..+2019^2 làm theo cách của lướp 6 nhé
Đáp án: $S=2^2+3^2+4^2+..+2019^2=\dfrac{2019.2020.2021.2022}{3}-\dfrac{2019.2020}{2}-1$ Giải thích các bước giải: $S=2^2+3^2+4^2+..+2019^2$ $\rightarrow S+1=1^2+2^2+3^2+4^2+..+2019^2$ $\rightarrow S+1=1.(2-1)+2.(3-1)+3.(4-1)+…+2019(2020-1)$ $\rightarrow S+1=(1.2+2.3+3.4+…+2019.2020)-(1+2+3+…+2019))$ Ta có: $A=1.2+2.3+3.4+..+2019.2020$ $\rightarrow 3A=1.2.3+2.3.3+..+2019.2020.3$ $\rightarrow 3A=1.2.(3-0)+2.3.(4-1)+..+2019.2020.(2021-2018)$ $\rightarrow 3A=1.2.3-0.1.2+2.3.4-1.2.3+..+2019.2020.2021-2018.2019.2020$ $\rightarrow 3A=2019.2020.2021$ $\rightarrow A=\dfrac{2019.2020.2021}{3}$ $\rightarrow S+1=\dfrac{2019.2020.2021}{3}-\dfrac{2019.2020}{2}$ $\rightarrow S=\dfrac{2019.2020.2021}{3}-\dfrac{2019.2020}{2}-1$ Bình luận
Đáp án:
$S=2^2+3^2+4^2+..+2019^2=\dfrac{2019.2020.2021.2022}{3}-\dfrac{2019.2020}{2}-1$
Giải thích các bước giải:
$S=2^2+3^2+4^2+..+2019^2$
$\rightarrow S+1=1^2+2^2+3^2+4^2+..+2019^2$
$\rightarrow S+1=1.(2-1)+2.(3-1)+3.(4-1)+…+2019(2020-1)$
$\rightarrow S+1=(1.2+2.3+3.4+…+2019.2020)-(1+2+3+…+2019))$
Ta có:
$A=1.2+2.3+3.4+..+2019.2020$
$\rightarrow 3A=1.2.3+2.3.3+..+2019.2020.3$
$\rightarrow 3A=1.2.(3-0)+2.3.(4-1)+..+2019.2020.(2021-2018)$
$\rightarrow 3A=1.2.3-0.1.2+2.3.4-1.2.3+..+2019.2020.2021-2018.2019.2020$
$\rightarrow 3A=2019.2020.2021$
$\rightarrow A=\dfrac{2019.2020.2021}{3}$
$\rightarrow S+1=\dfrac{2019.2020.2021}{3}-\dfrac{2019.2020}{2}$
$\rightarrow S=\dfrac{2019.2020.2021}{3}-\dfrac{2019.2020}{2}-1$