Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Lời giải:
\(B=\frac{5}{3}+\frac{5}{3^2}+\frac{5}{3^3}+...+\frac{5}{3^{30}}=5(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{30}})\)
\(3B=5(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{29}})\)
\(3B-B=5(1-\frac{1}{3^{30}})=\frac{5(3^{30}-1)}{3^{30}}\)
\(\Rightarrow B=\frac{5(3^{30}-1)}{2.3^{30}}\)
S = (-3)0 + (-3)1 + (-3)2 + ... + (-3)2015
=> 3S = (-3)1 + (-3)2 + (-3)3 + ... + (-3)2016
=> 3S + S = [(-3)1 + (-3)2 + ... + (-3)2016] + [(-3)0 + (-3)1 + ... + (-3)2015]
=> 4S = (-3)2016 + (-3)0
=> S = \(\frac{\left(-3\right)^{2016}+\left(-3\right)^0}{4}\)
\(a,A=2^0+2^1+2^2+....+\)\(2^{2010}\)
\(\Rightarrow2A=2^1+2^2+2^3+....+2^{2011}\)
\(2A-A=\left(2^1+2^2+2^3+...+2^{2011}\right)-\left(2^0+2^1+2^2+...+2^{2010}\right)\)
\(A=2^{2011}-2^0\)
\(A=2^{2011}-1\)
\(b,B=1+3+3^2+...+3^{100}\)
\(\Rightarrow3B=3+3^2+3^3+...+3^{101}\)
\(3B-B=\left(3+3^2+3^3+...+3^{101}\right)-\left(1+3+3^2+...+3^{100}\right)\)
\(2B=3^{101}-1\)
\(\Rightarrow B=\frac{3^{101}-1}{2}\)
\(c,C=4+4^2+4^3+...+4^n\)
\(\Rightarrow4C=4^2+4^3+4^4+...+4^{n+1}\)
\(4C-C=\left(4^2+4^3+4^4+...+4^{n+1}\right)-\left(4+4^2+4^3+...+4^n\right)\)
\(3C=4^{n+1}-4\)
\(\Rightarrow C=\frac{4^{n+1}-4}{3}\)
\(d,D=1+5+5^2+...+5^{2000}\)
\(\Rightarrow5D=5+5^2+5^3+...+5^{2001}\)
\(5D-D=\left(5+5^2+5^3+...+5^{2001}\right)-\left(1+5+5^2+...+5^{2000}\right)\)
\(4D=5^{2001}-1\)
\(\Rightarrow D=\frac{5^{2001}-1}{4}\)
b)
B=1+3+3^2+3^3+..+3^100
=> 3B = 3 + 3^2 + 3^3 + ...+ 3^101
=> 3B - B = ( 3 + 3^2 + 3^3 + ...+ 3^101) - (1+3+3^2+3^3+..+3^100)
=> 2B = 3^101 - 1
=> B =( 3^101 - 1) / 2
a) \(S=\left(-\frac{1}{7}\right)^0+\left(-\frac{1}{7}\right)^1+\left(-\frac{1}{7}\right)^2+...+\left(-\frac{1}{7}\right)^{2007}\)
\(=1+\left(-\frac{1}{7}\right)+\left(-\frac{1}{7}\right)^2+...+\left(-\frac{1}{7}\right)^{2007}\)
=> 7S = \(7+\left(-1\right)+\left(-\frac{1}{7}\right)+...+\left(-\frac{1}{7}\right)^{2006}\)
Lấy 7S trừ S ta có :
7S - S = \(7+\left(-1\right)+\left(-\frac{1}{7}\right)+...+\left(-\frac{1}{7}\right)^{2006}-\left[1+\left(-\frac{1}{7}\right)+\left(-\frac{1}{7}\right)^2+...+\left(-\frac{1}{7}\right)^{2007}\right]\)
6S = \(7-1-1+\left(\frac{1}{7}\right)^{2007}=5+\left(\frac{1}{7}\right)^{2007}\Rightarrow S=\frac{5+\left(\frac{1}{7}\right)^{2007}}{6}\)
Câu 2: Ta có \(S=6^2+18^2+30^2+...+126^2\)
\(S=6^2\left(1^2+3^2+5^2+...+21^2\right)\)
\(=6^2.1771=36.1771=63756\)
