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a) 3.32.33.....3100=31+2+...+100
b)x.x3.x5.....x49=x1+3+5+...+49
a, \(2+2^2+.....+2^{49}+2^{50}=2^{1+2+..+50}=2^{\frac{\left(50+1\right)\left[\left(50-1\right):1+1\right]}{2}}=1275\)
b, tương tự
a)\(\left(3^2+1\right)B=\left(3^2+1\right)\cdot3\cdot\left(1-3^2+3^4-3^6+3^8-...-3^{2006}+3^{2008}\right).\)
\(10B=3\cdot\left(3^{2010}+1\right)\)
\(B=\frac{3\left(3^{2010}+1\right)}{10}\)
b) \(B=3\cdot\left(1-3^2+3^4\right)-3^7\cdot\left(1-3^2+3^4\right)+...+3^{2005}\left(1-3^2+3^4\right)\)
\(B=\left(1-3^2+3^4\right)\cdot\left(3-3^7+3^{13}-...+3^{2005}\right)=73\cdot\left(3-3^7+3^{13}-...+3^{2005}\right)\)
chia hết cho 73.
a)B=3-3^3+3^5-3^7+3^9-...+3^2009
3^2B=3^3-3^5+3^7-3^9+3^11-...+3^2011
9B+B=3^3-3^5+3^7-3^9+3^11-...+3^2011+3-3^3+3^5-3^7+3^9-...+3^2009
10B=3^2011+3
B=\(\frac{3^{2011}+3}{10}\)
b) B=3-3^3+3^5-3^7+3^9-...+3^2009
=(3-3^3+3^5)-(3^7-3^9+3^11)-....+(3^2005-3^2007+3^2009)
=(3-3^3+3^5)-[3^6(3-3^3+3^5)]-...+[3^2004(3-3^3+3^5)]
=(3-3^3+3^5)-3^6(3-3^3+3^5)-...+3^2004(3-3^3+3^5)
=219(1-3^6-...+3^2004) chia hết cho 73 vì 219 chia hết cho 73
tớ nghĩ cùng mũ nhân tử : \(2^7.5^7=\left(2.5\right)^7=10^7\)
\(2^7.5^7=\left(2.5\right)^7=10^7\)