tính tổng sau S= 1. 2^2+ 2. 3^2+ 3. 4^2+……+2018 . 2019^2 làm theo cách lớp 6 nhé

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1. Đáp án:

    

   Giải thích các bước giải:

   S=1.$2^{2}$+2.$3^{2}$+3.$4^{2}$+…+2018.$2019^{2}$ 

   S=$\frac{2018 . 2019 . 2020 . 2021}{4}$ – $\frac{2018.2019.2020}{3}$ 

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2. Đáp án:

   $S=1.2^2+2.3^2+3.4^2+..+2018.2019^2=\dfrac{2018.2019.2020.2021}{4}-\dfrac{2018.2019.2020}{3}$

   Giải thích các bước giải:

   Ta có:

   $S=1.2^2+2.3^2+3.4^2+..+2018.2019^2$

   $\rightarrow S=1.2.2+2.3.3+3.4.4+…+2018.2019.2019$

   $\rightarrow S=1.2.(3-1)+2.3.(4-1)+..+2018.2019.(2020-1)$

   $\rightarrow S=1.2.3-1.2+2.3.4-2.3+…+2018.2019.2020-2018.2019$

   $\rightarrow S=1.2.3+2.3.4+3.4.5+..+2018.2019.2020-(1.2+2.3+3.4+..+2018.2019)$

   Ta có:

   $A=1.2.3+2.3.4+3.4.5+..+2018.2019.2020$

   $\rightarrow 4A=1.2.3.(4-0)+2.3.4.(5-1)+3.4.5.(6-2)+..+2018.2019.2020.(2021-2017)$

   $\rightarrow 4A=1.2.3.4-0.1.2.3.4+2.3.4.5-1.2.3.4+3.4.5.6-2.3.4.5+..+2018.2019.2020.2021-2017.2018.2019.2020$

   $\rightarrow 4A=2018.2019.2020.2021$

   $\rightarrow A=\dfrac{2018.2019.2020.2021}{4}$

   Lại có: 

   $B=1.2+2.3+3.4+..+2018.2019$

   $\rightarrow 3B=1.2.3+2.3.3+3.4.3+…+2018.2019.3$

   $\rightarrow 3B=1.2.(3-0)+2.3.(4-1)+3.4.(5-2)+…+2018.2019.(2020-2017)$

   $\rightarrow 3B=1.2.3-0.1.2+2.3.4-1.2.3+3.4.5-2.3.4+…+2018.2019.2020-2017.2018.2019$

   $\rightarrow 3B=2018.2019.2020$

   $\rightarrow B=\dfrac{2018.2019.2020}{3}$

   $\rightarrow S=A-B=\dfrac{2018.2019.2020.2021}{4}-\dfrac{2018.2019.2020}{3}$

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