Rút gọn tổng sau:
M=1-2+2^2-2^3+ ...+2^2008
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\(M=1-2+2^2-2^3+...+2^{2008}\)
\(\Rightarrow2M=2-2^2+2^3-2^4+...-2^{2008}+2^{2009}\)
\(\Rightarrow2M+M=2-2^2+2^3-2^4+...-2^{2008}+2^{2009}+1-2+2^2-2^3+...+2^{2008}\)
\(3M=2^{2009}+1\)
\(\Rightarrow M=\frac{2^{2009}+1}{3}\)
cop bài trên mạng ko có càu này mik làm r rồi chụp lại nhé ;-;
a)
`A = 1 + 2 + 2^2 + .....+2^2015`
=>`2A = 2 + 2^2 + 2^3 + ... + 2^2016`
=> `2A - A= (2 + 2^2 + 2^3 + ... + 2^2016)-(1 + 2 + 2^2 + ... + 2^2015)`
=> `A = 2^2016 - 1`
b) `4^2008 = (2^2)^2008 = 2^4016 > 2^2016 - 1`
đặt tử là A,ta có:
2A=2(1+2+22+23+...+22008)
2A=2*1+2*2+2*22+...+2*22008
2A=2+22+23+...+22009
2A-A=(2+22+23+...+22009)-(1+2+22+...+22008)
A=22009-1
thay A vào tử số ta được \(B=\frac{2^{2009}-1}{1-2^{2009}}=-1\)
\(A=2^0+2^1+2^2\)\(+2^3+...+\)\(2^{50}\)
\(2A=2+2^2+2^3+...+2^{51}\)
\(2A-A=A=2^{51}-2^0\)
\(B=5+5^2+5^3+...+5^{99}+5^{100}\)
\(5B=5^2+5^3+5^4+...+5^{100}+5^{101}\)
\(5B-B=4B=5^{101}-5\)
\(B=\frac{5^{101}-5}{4}\)
\(C=3-3^2+3^3-3^4+...+\)\(3^{2007}-3^{2008}+3^{2009}-3^{2010}\)
\(3C=3^2-3^3+3^4-3^5+...-3^{2008}+3^{2009}-3^{2010}+3^{2011}\)
\(3C+C=4C=3^{2011}+3\)
\(C=\frac{3^{2011}+3}{4}\)
\(S_{100}=5+5\times9+5\times9^2+5\times9^3+...+5\times9^{99}\)
\(S_{100}=5\times\left(1+9+9^2+9^3+...+9^{99}\right)\)
\(9S_{100}=5\times\left(9+9^2+9^3+...+9^{99}+9^{100}\right)\)
\(9S_{100}-S_{100}=8S_{100}=5\times\left(9^{100}-1\right)\)
\(S_{100}=\frac{5\times\left(9^{100}-1\right)}{8}\)
A=20+21+22+23+...++23+...+250250
2�=2+22+23+...+2512A=2+22+23+...+251
2�−�=�=251−202A−A=A=251−20
�=5+52+53+...+599+5100B=5+52+53+...+599+5100
5�=52+53+54+...+5100+51015B=52+53+54+...+5100+5101
5�−�=4�=5101−55B−B=4B=5101−5
�=5101−54B=45101−5
�=3−32+33−34+...+C=3−32+33−34+...+32007−32008+32009−3201032007−32008+32009−32010
3�=32−33+34−35+...−32008+32009−32010+320113C=32−33+34−35+...−32008+32009−32010+32011
3�+�=4�=32011+33C+C=4C=32011+3
�=32011+34C=432011+3
�100=5+5×9+5×92+5×93+...+5×999S100=5+5×9+5×92+5×93+...+5×999
�100=5×(1+9+92+93+...+999)S100=5×(1+9+92+93+...+999)
9�100=5×(9+92+93+...+999+9100)9S100=5×(9+92+93+...+999+9100)
9�100−�100=8�100=5×(9100−1)9S100−S100=8S100=5×(9100−1)
�100=5×(9100−1)8S100=85×(9100−1)
Đặt biểu thức là A:
\(A=-6.2009^2-2^2.2009=-6.2007.2009.2011\)
\(A=\frac{-6.2009}{2005.2010}\)
\(A=\frac{-2009}{2005.335}\)
P/s: Ko chắc
Lời giải:
$M=1-2+2^2-2^3+...+2^{2008}$
$2M=2-2^2+2^3-2^4+...+2^{2009}$
$M+2M=(1-2+2^2-2^3+...+2^{2008})+(2-2^2+2^3-2^4+...+2^{2009})$
$3M=2^{2009}+1$
$M=\frac{2^{2009}+1}{3}$