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a) \(M=1+3+3^2+3^3+...+3^{119}\)
\(3M=3+3^2+3^3+3^4+...+3^{119}+3^{120}\)
\(3M-M=\left(3+3^2+3^3+...+3^{120}\right)-\left(1+3+3^2+...+3^{119}\right)\)
\(2M=3^{120}-1\)
\(M=\frac{3^{120}-1}{2}\)
b) \(M=1+3+3^2+3^3+...+3^{118}+3^{119}\)
\(=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+...+\left(3^{117}+3^{118}+3^{119}\right)\)
\(=\left(1+3+3^2\right)+3^3\left(1+3+3^2\right)+...+3^{117}\left(1+3+3^2\right)\)
\(=13\left(1+3^3+...+3^{117}\right)\)chia hết cho \(13\).
\(M=1+3+3^2+3^3+...+3^{118}+3^{119}\)
\(=\left(1+3+3^2+3^3\right)+...+\left(3^{116}+3^{117}+3^{118}+3^{119}\right)\)
\(=\left(1+3+3^2+3^3\right)+...+3^{116}\left(1+3+3^2+3^3\right)\)
\(=40\left(1+3^4+...+3^{116}\right)\)chia hết cho \(5\).
\(3^1+3^2+3^3+3^4+3^5+...+\)\(3^{2012}\)
\(=(3^1+3^2+3^3+3^4)+(3^5+3^6+3^7+3^8)+...+\)\((\)\(3^{2009}\)\(+\)\(3^{2010}\)\(+\)\(3^{2011}\)\(+\)\(3^{2012}\)\()\)
\(=1(3^1+3^2+3^3+3^4)+4(3^1+3^2+3^3+3^4)+...+2008(3^1+3^2+3^3+3^4)\)
\(=(1+4+...+2008). (3^1+3^2+3^3+3^4)\)
\(=Q.120\)
\(\Rightarrow\) Tổng \(3^1+3^2+3^3+3^4+3^5+...+\)\(3^{2012}\) \(⋮\) \(120\)
31 + 32 + 33+ 34 + 35 + … + 32012
= (31 + 32 + 33+ 34) + (35 + 36 + 37 + 38) + ... + (32009 + 32010 + 32011 + 32012)
= 1(31 + 32 + 33+ 34) + 34(31 + 32 + 33+ 34) + ... + 32008(31 + 32 + 33+ 34)
= (1 . 120) + (34 . 120) + ... + (32008 . 120)
= (1 + 34 + ... + 32008) . 120
= 120 ⋮ 120
⇒ Tổng 31 + 32 + 33+ 34 + 35 + … + 32012 chia hết cho 120
2^n . 4^2 - 2^n + 1 = 2^6 - 2^3
2^n . 4^2 - 2^n . 1 + 1 = 2^6 - 2^3
2^n . (4^2 - 1) + 1 = 2^3 . 2^3 - 2^3 . 1
2^n . (16 - 1) + 1 = 2^3 . (2^3 - 1)
2^n . 15 + 1 = 8 . (8 - 1)
2^n . 15 + 1 = 8 . 7
2^n . 15 + 1 = 56
2^n . 15 = 56 - 1
2^n . 15 = 55
2^n = 55 : 15
2^n = 11/3
=> Không tồn tại n
Mình nghĩ là vậy, không biết đúng không. Nếu sai thì sorry nha! ^_^
2^n.4^2-2^n+1=15.2^n+1
15.2^n+1=2^3.7
2^n+4 -2^n+1=56
2^n+4-2^n-55=0
suy ra :n=3749 phần 2000 (viết theo dạng phân số)
7x + 135 : 45=52
7x + 3 = 52
7x= 52 - 3
7x= 49
7x= \(7^2\)
\(\Rightarrow\) x=2