( 1.2 + 2.3 + 3.4 + ....... + 2012.2013) - ( 22 + 32 + 42 + 52 ........ + 20132 )
12 /1.2 . 22 /2.3 . 32 /3.4 . 42 /4.5 .52 /5.6 làm thế nào mọi người giúp với ạ
Tính C = 1 2 – 2 2 + 3 2 – 4 2 + 5 2 – 6 2 + … . + 2013 2 – 2014 2 + 2015 2
C=12-22+32-42+52-62+..+20132-20142+20152
SSH:(20152-12):10+1=2015
(12-22)+(32-42)+(52-62)+...+(20132-20142)+20152
-10+(-10)+(-10)+...+(-10)+20152
-10x(2015-1):2+20152=12
=> C=12
tính tổng M=1.2+2.3+3.4+........+2012.2013
Ta có:
3M=1.2.3+2.3.3+3.4.3+...+2012.2013.3
3M=1.2.3+2.3.(4-1)+3.4.(5-2)+...+2012.2013.(2014-2011)
3M=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+2012.2013.2014-2011.2012.2013
3M=2012.2013.2014
3M=8157014184
M=8157014184:3
M=2719004728
Tính (1.2+2.3+3.4+...2012.2013)-(2^2+3^2+...+2013^2)
Lời giải:
$(1.2+2.3+3.4+...+2012.2013)-(2^2+3^2+...+2013^2)$
$=[(2-1).2+(3-1).3+(4-1).4+...+(2013-1).2013]-(2^2+3^2+...+2013^2)$
$=(2^2+3^2+4^2+...+2013^2)-(2+3+4+...+2013)-(2^2+3^2+...+2013^2)$
$=-(2+3+4+...+2013)$
$=1-(1+2+3+...+2013)$
$=1-2013.2014:2=1-2027091=-2027090$
Tính nhanh:
(1.2+2.3+3.4+....+2012.2013)-(2.2+3.3+4.4+...+2013.2103)
Tín (1.2+2.3+3.4+4.5+.......+2012.2013)-(2.2+3.3+4.4+5.5+.......+2013.2013)
(1.2+2.3+3.4+.....+2012.2013)-(22+32+42+......+20132)
Tính nhanh
(1.2+2.3+3.4+.....+2012.2013)-(22+32+42+......+20132)
= 1.2 + 2.3 + 3.4 +...+ 2012.2013 - 22 -32 - 42 -....-20132
=1.2 + 2.3 + 3.4 + ...+2012.2013 - 2.2 -3.3 - 4.4 -...- 2013.2013
=(1.2 - 2.2) + (2.3 - 3.3) + (3.4 - 4.4) + ...+(2012.2013 - 2013.2013)
=2.(1-2) + 3.(2-3) + 4.(3-4) +...+2013.(2012-2013)
=2.(-1) + 3.(-1) + 4.(-1) + ...+2013.(2012-2013)
= -2 - 3 - 4 -...- 2013
= -(2+3+4+...+2013)
= -[(2013+2).2012:2]
=-2027090
Bài 1
A=1.2+2.3+3.4+....+151.152
B=1.3+3.5+5.7+...+2023.2025
C=2.4+4.6+...+2024.2026
D=1.2+3.4+...+200.202
M=12+22+...+20242
N=13+23+...+1003
Q=13+23+...+20243
R=12+22+...+2003
\(A=1\cdot2+2\cdot3+...+151\cdot152\)
\(=1\left(1+1\right)+2\left(1+2\right)+...+151\left(1+151\right)\)
\(=\left(1+2+3+...+151\right)+\left(1^2+2^2+...+151^2\right)\)
\(=\dfrac{151\left(151+1\right)}{2}+\dfrac{151\left(151+1\right)\left(2\cdot151+1\right)}{6}\)
\(=151\cdot76+\dfrac{151\cdot152\cdot303}{6}\)
\(=151\cdot76+151\cdot7676=1170552\)
\(C=2\cdot4+4\cdot6+...+2024\cdot2026\)
\(=2\cdot2\left(1\cdot2+2\cdot3+...+1012\cdot1013\right)\)
\(=4\left[1\left(1+1\right)+2\left(1+2\right)+...+1012\left(1+1012\right)\right]\)
\(=4\left[\left(1+2+...+1012\right)+\left(1^2+2^2+...+1012^2\right)\right]\)
\(=4\left[1012\cdot\dfrac{1013}{2}+\dfrac{1012\left(1012+1\right)\left(2\cdot1012+1\right)}{6}\right]\)
\(=4\left[506\cdot1013+345990150\right]\)
\(=1386010912\)
\(M=1^2+2^2+...+2024^2\)
\(=\dfrac{2024\left(2024+1\right)\cdot\left(2\cdot2024+1\right)}{6}\)
\(=2024\cdot2025\cdot\dfrac{4049}{6}\)
=2765871900
\(N=1^3+2^3+...+100^3\)
\(=\left(1+2+3+...+100\right)^2\)
\(=\left[\dfrac{100\left(100+1\right)}{2}\right]^2\)
\(=\left[50\cdot101\right]^2=5050^2\)
\(Q=1^3+2^3+...+2024^3\)
\(=\left(1+2+3+...+2024\right)^2\)
\(=\left[\dfrac{2024\left(2024+1\right)}{2}\right]^2\)
\(=\left[1012\left(2024+1\right)\right]^2\)
\(=2049300^2\)