Cho A= 1/2+ 3/2 + (3/2)2+ (3/2)3+......+(3/2)2012
B= (3/2)2012:2
Tính A-B
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\(\frac{3}{2}.A=\frac{3}{4}+\left(\frac{3}{2}\right)^2+\left(\frac{3}{2}\right)^3+...+\left(\frac{3}{2}\right)^{2013}\)
\(\Rightarrow\frac{3}{2}.A-A=\frac{3}{4}+\left(\frac{3}{2}\right)^2+\left(\frac{3}{2}\right)^3+...+\left(\frac{3}{2}\right)^{2013}-\left(\frac{1}{2}+\frac{3}{2}+\left(\frac{3}{2}\right)^2+...+\left(\frac{3}{2}\right)^{2012}\right)\)
\(\Rightarrow\frac{1}{2}.A=\frac{3}{4}+\left(\frac{3}{2}\right)^{2013}-\frac{1}{2}-\frac{3}{2}=\left(\frac{3}{2}\right)^{2013}-\frac{5}{4}\Rightarrow A=2.\left(\frac{3}{2}\right)^{2013}-\frac{5}{2}\)
\(B-A=\frac{1}{2}.\left(\frac{3}{2}\right)^{2013}-2.\left(\frac{3}{2}\right)^{2013}+\frac{5}{2}=-\left(\frac{3}{2}\right)^{2014}+\frac{5}{2}\)
Xét biểu thức A
A= 1+(1+2) +....... +(1+2+3+...+2012)
A = 1+1+2+1+2+3+...+1+2+3+...+2012
A có 2012 số 1
có 2011 số 2
...
có 1 số 2012
A = 1 x2012 +2x2011+...+2012x1
mà B = 1 x2012 +2x2011+...+2012x1
nên A=B
\(A=1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+2012\right)\)
\(=\left(1+1+1+...+1\right)+\left(2+2+...+2\right)+...+2012\)
\(=1\times2012+2\times2011+...+2012\times1\)
\(=B\)
c) Cho B = (1.2.3....2012) . ( 1 + \(\dfrac{1}{2}\) + \(\dfrac{1}{3}\) + ... + \(\dfrac{1}{2012}\) ) Chứng minh B chia hết cho 2013
B = (1.2.3....2012) . (1 + \(\dfrac{1}{2}\) + \(\dfrac{1}{3}\) + ...+ \(\dfrac{1}{2012}\) )
=(1.2.3...671...2012) . (1 + \(\dfrac{1}{2}\) + \(\dfrac{1}{3}\) + ... + \(\dfrac{1}{2012}\))
=(1.2.(3.671)...2012) . (1 + \(\dfrac{1}{2}\) +\(\dfrac{1}{3}\) + ... + \(\dfrac{1}{2012}\))
=(1.2.2013...2012) . (1 + \(\dfrac{1}{2}\) + \(\dfrac{1}{3}\) + ... + \(\dfrac{1}{2012}\))
Vậy B chia hết cho 2013
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Lời giải:
$A-\frac{1}{2}=\frac{3}{2}+(\frac{3}{2})^2+....+(\frac{3}{2})^{2012}$
$\frac{3}{2}(A-\frac{1}{2})=(\frac{3}{2})^2+(\frac{3}{2})^3+....+(\frac{3}{2})^{2013}$
$\Rightarrow \frac{3}{2}(A-\frac{1}{2}) - (A-\frac{1}{2})=(\frac{3}{2})^{2013}-\frac{3}{2}$
$\Rightarrow \frac{1}{2}(A-\frac{1}{2})=(\frac{3}{2})^{2013}-\frac{3}{2}$
$\Rightarrow A=2(\frac{3}{2})^{2013}-\frac{5}{2}$
$\Rightarrow A-B=2(\frac{3}{2})^{2013}-\frac{5}{2}- \frac{1}{2}.(\frac{3}{2})^{2013}$
$\Rightarrow A-B=\frac{3}{2}(\frac{3}{2})^{2013}-\frac{5}{2}=(\frac{3}{2})^{2014}-\frac{5}{2}$
Trả lời:
Ta có:
A= \(\frac{1}{2}\)+\(\frac{3}{2}\)+(\(\frac{3}{2}\))2+...+(\(\frac{3}{2}\))2012 (1)
\(\frac{3}{2}\)A= \(\frac{3}{4}\)+(\(\frac{3}{2}\))2+.....+(\(\frac{3}{2}\))2013 (2)
Lấy (2)-(1)
=> \(\frac{1}{2}\)A=(\(\frac{3}{2}\)) 2013 -\(\frac{5}{4}\)
=> A=(\(\frac{3^{2013}}{2^{2012}}\) - \(\frac{5}{2}\))
B=( \(\frac{3}{2}\))2013 :2= \(\frac{3^{2013}}{2^{2014}}\)
=> A-B= (\(\frac{3^{2013}}{2^{2012}}\) - \(\frac{5}{2}\))- \(\frac{3^{2013}}{2^{2014}}\)