a) 1+2+2^2+2^3+2^4+...+2^x-1=2^95-1.
b) 1+2+2^2+2^3+2^4+...+2^x=2^75–1.
c) 2^x-1+2^x+2^x+1+2^x+2+2^x+3=992.
d) 2+2•3+2•3^2+2•3^3+...+2•3^2011=x•81^502–1
e) 2+2•3+2•3^2+2•3^3+...+2•3^2013=x•729•81^501–1.
f) 2+2^2+2^3+2^4+...+2^2013+2^2014=16(8x+1)•32^401–4•8^669–2.
DAU COROANA VA ROG REPEDE
Răspunsuri la întrebare
Răspuns:
Fie S = 2^0 + 2^1 + 2^2 + ... +2^n
2S = 2^1 + 2^2 + 2^3 + ... + 2^(n+1)
S = 2S - S
2^1 se reduce
2^2 se reduce
..............
2^n se reduce
Ramanem S = 2^(n+1) - 1
a) 2^(x-1+1) - 1 = 2^95 - 1 |+1
2^x = 2^95
x = 95
b) 2^(x+1) - 1 = 2^75 - 1 |+1
2^(x+1) = 2^74+1
x = 74
c) Generalizam,
S = 2^n + 2^n+1 + 2^n+2 + ... + 2^x
S =
= 2S - S =
= 2^x+1 - 2^n
Trecem la problema noastra.
2^x+4 - 2^x-1 = 992
2^x-1 (2^5 - 1) = 992
2^x-1 (32-1) = 992
2^x-1 = 992/31
2^x-1 = 32
2^x-1 = 2^6-1
x = 6
d) 2(3^0 + 3^1 + 3^2 + ... + 3^2011) = x*81^502 - 1
S = 3^0 + 3^1 + ... + 3^n
3S = 3^1 + 3^2 + ... + 3^n+1
2S =
= 3S - S =
= 3^n+1 - 1
S = (3^n+1 - 1)/2
x*81^502 - 1 = 2*(3^2012-1)/2
x*81^502 - 1 = 3^2012 - 1 |+1
x*3^2008 = 3^2012
x = 3^2012 / 3^2008
x = 3^4
x = 81
e) 2*(3^2014 - 1)/2 = x*729*81^501 - 1
x*3^6*3^2004 - 1 = 3^2014 - 1 |+1
x*3^2010 = 3^2014
x = 3^2014/3^2010
x = 3^4
x = 81
f) 2^2015 - 1 = 16(8x+1)*32^401 - 4*8^669 - 2
Rezolvam partea dreapta.
16 = 2^4
32^401 = (2^5)^401 = 2^2005
2^4*2^2005 = 2^2009
Avem 2^2009(8x+1)
4 = 2^2
8^669 = (2^3)^669 = 2^2007
2^2 * 2^2007 = 2^2009
Avem 2^2009
Ramanem cu 2^2009(8x+1 - 1) - 2 =
= 2^2009*8x - 2 =
= 2^2012*x - 2 =
= 2(2^2011*x - 1)
Rescriem ecuatia :
2(2^2011*x - 1) = 2^2015 - 1 |:2
2^2011*x - 1 = 2^2014 - 1/2 |+1
2^2011*x = 2^2014 + 1/2
x = (2^2014 + 1/2) / 2^2011
x = 2^3 + 1/2^2012
x = 2^2015/2^2012 + 1/2^2012
x = (2^2015 + 1) / 2^2012
M.am chinuit ceva sa le fac, sper ca n.am gresit nimic :)