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\(1^3=0+1\)
\(2^3=2+1.2.3\)
.......................................
\(10^3=10+9.10.11\)
\(=1+2+1.2.3+3+2.3.4+....+10+9.10.11\)
\(=\left(1+2+....+10\right)+\left(1.2.3+....+9.10.11\right)\)
Đặt (1 + 2 + ... + 10) là A ; (1.2.3 + 2 .3.4 + .... + 9.10.11) là B . Ta có :
\(A=\left(1+2+...+10\right)\)
\(A=\frac{\left(10+1\right).10}{2}\)
\(A=55\)
\(B=1.2.3+2.3.4+....+9.10.11\)
\(4B=1.2.3.4+2.3.4.4+...+9.10.11.4\)
\(4B=1.2.3.\left(4-0\right)+2.3.4.\left(5-1\right)+....+9.10.11.\left(12-8\right)\)
\(4B=1.2.3.4+2.3.4.5+....+9.10.11.12\)
\(4B=9.10.11.12=11880\)
\(\Rightarrow B=\frac{11880}{4}=2970\)
\(\Rightarrow1^3+2^3+....+10^3=A+B=55+2970=3025\)
Ta có:A=\(1+3+3^2+3^3+...+3^{2012}\)
3A=\(3\cdot\left(1+3+3^2+3^3+...+3^{2012}\right)\)
3A=\(3+3^2+3^3+3^4+...+3^{2013}\)
3A-A=\(\left(3+3^2+3^3+3^4+...+3^{2013}\right)-\left(1+3+3^2+3^3+...+3^{2012}\right)\)
2A=\(3+3^2+3^3+3^4+...+3^{2013}-1-3-3^2-3^3-...-3^{2012}\)
2A=\(\left(3-3\right)+\left(3^2-3^2\right)+\left(3^3-3^3\right)+...+\left(3^{2012}-3^{2012}\right)+\left(3^{2013}-1\right)\)
2A=\(0+0+0+...+0+3^{2013}-1\)
2A=\(3^{2013}-1\)
A=\(\frac{3^{2013}-1}{2}\)
B=\(3^{2013}\div2\)
B=\(\frac{3^{2013}}{2}\)
VậyB-A=\(\frac{3^{2013}}{2}-\frac{3^{2013}-1}{2}\)
\(B-A=\frac{3^{2013}-\left(3^{2013}-1\right)}{2}\)
\(B-A=\frac{3^{2013}-3^{2013}+1}{2}\)
\(B-A=\frac{1}{2}=0,5\)
\(S=6^2\left(1^2+2^2+3^2+...+10^2\right)=36.385=13860\)
\(A=1+2^2+2^3+...+2^{2018}\)
\(2A=2+2^2+...+2^{2019}\)
\(2A-A=\left(2+2^2+...+2^{2019}\right)-\left(1+2^2+2^3+...+2^{2018}\right)\)
\(A=2^{2019}-1\)
\(\Rightarrow A+1=2^{2019}-1+1=2^{2019}\)
\(\Rightarrow A+1\)là một lũy thừa
đpcm
Tính:
(-2)2.3 -(110+8):(-3)2
=4.3-(1+8):9
=12-9:9
=12-1
=11