8 Tính nhanh:
I=\(^{1^2+2^2+3^2+...+2017^2}\)
K
Khách
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Những câu hỏi liên quan
11 tháng 8 2020
\(I=1\left(2-1\right)+2\left(3-1\right)+...+2017\left(2018-1\right)\)
\(I=\left(1.2+2.3+...+2017.2018\right)-\left(1+2+...+2017\right)\)
\(3I=\left(1.2.3+2.3\left(4-1\right)+...+2017.2018\left(2019-2016\right)\right)-\frac{3.2017.2018}{2}\)
=> \(I=\frac{2017.2018.2019}{3}-2017.1009\)
=> \(I=.....\)
DD
1
OT
18 tháng 4 2016
999 - 888 - 111 + 111 - 111 + 111 - 111
= 111 - 111 + 111 - 111 + 111 - 111
= 0 + 111 - 111 + 111 - 111
= 111 - 111 + 111 - 111
= 0 + 111 - 111
= 111 - 111
= 0
TL
3
DP
3 tháng 7 2017
a, \(S_1=3+4+6+8+...+2016+2017\)
\(S_1=3+\left(4+6+8+...+2016\right)+2017\)
Số số hạng của (4 + 6 + 8 + ... + 2016) là:
\(\left(2016-4\right)\div2+1=1007\)
Tổng của (4 + 6 + 8+ ... + 2016) là:
\(\frac{\left(4+2016\right).1007}{2}=1017070\)
\(\Rightarrow S_1=3+4+6+8+..+2016+2017=3+1017070+2017=1019090\)
b, \(S_2=2+3+5+7+...+2017+2018\)
\(S_2=2+\left(3+5+7+...+2017\right)+2018\)
Số số hạng của (3 + 5 + 7 + ... + 2017) là:
\(\frac{2017-3}{2}+1=1008\)
Tổng của (3 + 5 + 7 + ... + 2017) là:
\(\frac{\left(3+2017\right).1008}{2}=1018080\)
\(\Rightarrow S_2=2+3+5+7+...+2017+2018=2+1018080+2018=1020100\)
5 tháng 11 2021
\(a,2017\times\left(8+2\right)=2017\times10=20170\\ b,=\dfrac{3}{7}\times\left(\dfrac{9}{7}-\dfrac{2}{7}\right)=\dfrac{3}{7}\times1=\dfrac{3}{7}\)
5 tháng 11 2021
\(a,P=\dfrac{1}{\left(2+1\right)\left(2+1-1\right):2}+\dfrac{1}{\left(3+1\right)\left(3+1-1\right):2}+...+\dfrac{1}{\left(2017+1\right)\left(2017+1-1\right):2}\\ P=\dfrac{1}{2\cdot3:2}+\dfrac{1}{3\cdot4:2}+...+\dfrac{1}{2017\cdot2018:2}\\ P=2\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2017\cdot2018}\right)\\ P=2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2017}-\dfrac{1}{2018}\right)\\ P=2\left(\dfrac{1}{2}-\dfrac{1}{2018}\right)=2\cdot\dfrac{504}{1009}=\dfrac{1008}{1009}\)
\(b,\) Ta có \(\dfrac{1}{4^2}< \dfrac{1}{2\cdot4};\dfrac{1}{6^2}< \dfrac{1}{4\cdot6};...;\dfrac{1}{\left(2n\right)^2}< \dfrac{1}{\left(2n-2\right)2n}\)
\(\Leftrightarrow VT< \dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+...+\dfrac{1}{\left(2n-2\right)2n}\\ \Leftrightarrow VT< \dfrac{1}{2}\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+...+\dfrac{2}{\left(2n-2\right)2n}\right)\\ \Leftrightarrow VT< \dfrac{1}{2}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2n-2}-\dfrac{1}{2n}\right)\\ \Leftrightarrow VT< \dfrac{1}{2}\left(1-\dfrac{1}{2n}\right)< \dfrac{1}{2}\cdot\dfrac{1}{2}=\dfrac{1}{4}\)
KP
0
\(I=1^2+2^2+3^2+4^2+...+2017^2+2018^2\)
\(I=\left(2^2-1^2\right)+\left(4^2-3^2\right)+...+\left(2018^2-2017^2\right)\)
\(I=\left(1+2\right)\left(2-1\right)+\left(3+4\right)\left(4-3\right)+...+\left(2017+2018\right)\left(2018-2017\right)\)
\(I=1+2+3+4+...+2017+2018\)
\(I=\frac{\left(2018+1\right).2018}{2}=2037171\)
I = 12 + 22 + 32 + ... + 20172
= 1.1 + 2.2 + 3.3 + ... + 2017.2017
= 1.(2 - 1) + 2.(3 - 1) + 3.(4 - 1) + .... + 2017.(2018 - 1)
= 1.2 + 2.3 + 3.4 + ... + 2017.2018 - (1 + 2 + 3 + 4 + ... + 2017)
= 1.2 + 2.3 + 3.4 + ... + 2017.2018 - 2035153
Đặt K = 1.2 + 2.3 + 3.4 + ... + 2017.2018
=> 3K = 1.2.3 + 2.3.3 + 3.4.3 + .... + 2017.2018.3
=> 3K = 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 2017.2018.(2019 - 2016)
=> 3K = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + .... + 2017.2018.2019 - 2016.2017.2018
=> 3K = 2017.2018.2019
=> K = 2017.2018.2019 : 3
=> K = 2739315938
Lại có I = K - 2035153
= 2739315938 - 2035153
= 2 737 280 785