các bạn giúp mik với
2016/2+2016/6+2016/12+........+2016/9900
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Ta có :
2016 - 2013 = 3
2010 - 2007 = 3
Có số lần như vậy :
( 2016 - 3 ) : 3 + 1 = 672
Vì cứ 2 số trong dãy trừ nhau bằng 2 nên có :
Tổng của dãy trên :
672 x 3 = 2016
Đ/s : 2016
\(P=\dfrac{3^{2016}-6^{2016}+9^{2016}-12^{2016}+15^{2016}-18^{2016}}{-1^{2016}+2^{2016}-3^{2016}+4^{2016}-5^{2016}+6^{2016}}\)
\(=\dfrac{\left(3^{2016}-6^{2016}\right)+\left(9^{2016}-12^{2016}\right)+\left(15^{2016}-18^{2016}\right)}{-1^{2016}+2^{2016}-3^{2016}+4^{2016}-5^{2016}+6^{2016}}\)
\(=\dfrac{3^{2016}\left(1-2^{2016}\right)+3^{2016}\left(3^{2016}-4^{2016}\right)+3^{2016}\left(5^{2016}-6^{2016}\right)}{-1^{2016}+2^{2016}-3^{2016}+4^{2016}-5^{2016}+6^{2016}}\)
\(=\dfrac{3^{2016}\left(1-2^{2016}+3^{2016}-4^{2016}+5^{2016}-6^{2016}\right)}{-\left(1^{2016}-2^{2016}+3^{2016}-4^{2016}+5^{2016}-6^{2016}\right)}\)
\(=-3^{2016}\).
Vậy \(P=-3^{2016}\)
\(P=\frac{3^{2016}-6^{2016}+9^{2016}-12^{2016}+15^{2016}-18^{2016}}{-1^{2016}+2^{2016}-3^{2016}+4^{2016}-5^{2016}+6^{2016}}\)
\(=\frac{\left(1.3\right)^{2016}-\left(2.3\right)^{2016}+\left(3.3\right)^{2016}-\left(4.3\right)^{2016}+\left(5.3\right)^{2016}-\left(6.3\right)^{2016}}{-1^{2016}+2^{2016}-3^{2016}+4^{2016}-5^{2016}+6^{2016}}\)
\(=\frac{1^{2016}.3^{2016}-2^{2016}.3^{2016}+3^{2016}.3^{2016}-4^{2016}.3^{2016}+5^{2016}.3^{2016}-6^{2016}.3^{2016}}{-1^{2016}+2^{2016}-3^{2016}+4^{2016}-5^{2016}+6^{2016}}\)
\(=\frac{-3^{2016}\left(-1^{2016}+2^{2016}-3^{2016}+4^{2016}-5^{2016}+6^{2016}\right)}{-1^{2016}+2^{2016}-3^{2016}+4^{2016}-5^{2016}+6^{2016}}\)
\(=-3^{2016}\)
\(\frac{2016^{26}+2016^{24}+...+2016^4+2016^2}{2016^{24}+2016^{22}+...+2016^2+1}\) \(=\frac{2016^2.\left(2016^{24}+2016^{22}+...+2016^2+1\right)}{2016^{24}+2016^{22}+...+2016^2+1}\)
\(=\frac{2016^2}{1}=2016^2\)
=2016x(1/1x2+1/2x3+...+1/99x100)
=2016x(1-1/100)
=2016x99/100
=199584/100