Help me do my homework !
Cho \(A=1+3+3^2+3^3+3^4+.....+3^{2012}\) và \(B=3^{2013}:2\)
Tính B - A
Can you help me ?
K
Khách
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Sao giống lớp 6 quá nên mình trả lời luôn
Ta có: 3A=3+\(^{3^2+3^3+3^4+3^5+...+3^{2012}+3^{2013}}\)
\(\Rightarrow\)3A-A=2A=(\(3+3^2+3^3+3^4+...+3^{2013}\)) - (\(1-3^{ }-3^2-3^3-3^4-...-3^{2012}\))
\(\Rightarrow\)2A=\(3^{2013}-1\)\(\Rightarrow\)A=\(\left(3^{2013}-1\right):2\)\(\Rightarrow\)B-A=(\(^{\left(3^{2013}:2\right)-\left(\left(3^{2013}-1\right):2\right)\Rightarrow}\)
A = 1 + 3 + 32 +...+ 32012
3A = 3 + 32 + 33 +...+ 32013
3A - A = (3 + 32 + 33 +...+ 32013) - (1 + 3 + 32 +...+ 32012)
2A = 32013 - 1
A = \(\frac{3^{2013}-1}{2}\)
=> B - A = \(\frac{3^{2013}}{2}-\frac{3^{2013}-1}{2}=\frac{3^{2013}-\left(3^{2013}-1\right)}{2}=\frac{3^{2013}-3^{2013}+1}{2}=\frac{1}{2}\)