bạn nhóm 3 số vào 1 nhóm rồi nhóm ts chung riêng nhóm thứ nhất tính ra lun
Giải
Ta có: S=\(3^0+3^2+3^4+...+3^{2002}\)
\(\Leftrightarrow\)\(3^2\)S=\(3^2\)(\(3^0+3^2+3^4+...+3^{2002}\))
\(\Leftrightarrow\)\(3^2S=3^2+3^4+3^6+...+3^{2004}\)
\(\)\(3^2S-S=\left(3^2+3^4+3^6+...+3^{2004}\right)-\left(3^0+3^2+3^4+...+3^{2002}\right)\)
8S=\(\left(3^2-3^2\right)+\left(3^4-3^4\right)+\left(3^6-3^6\right)+...+\left(3^{2002}-3^{2002}\right)+3^{2004}-1\)
8S=0+0+0+...+\(3^{2004}\)-1=\(3^{2004}-1\)
\(\Leftrightarrow\)S=\(\frac{3^{2004}-1}{8}\)
\(S=3^0+3^2+3^4+...+3^{2002}\)
\(9S=3^2+3^4+3^6+...+3^{2004}\)
\(9S-S=\left(3^2+3^4+3^6+...+3^{2004}\right)-\left(3^0+3^2+3^4+...+3^{2002}\right)\)
\(9S-S=3^2+3^4+3^6+...+3^{2004}-3^0-3^2-...-3^{2002}\)( ÁP DỤNG CÔNG THỨC DẤU NGOẶC)
\(8S=3^{2004}-3^0\)(TRỪ ĐI CÁC LŨY THỪA GIỐNG NHAU)
\(S=3^{2004}-3^0\div8\)
b) \(S=\left(3^0+3^2+3^4\right)+\left(3^6+3^8+3^{10}\right)+...+\left(3^{1998}+3^{2000}+3^{2002}\right)\)
\(S=3^0\left(1+3^2+3^4\right)+3^6\left(1+3^2+3^4\right)+...+3^{1998}\left(1+3^2+3^6\right)\)
\(S=3^0.91+3^6.91+...+3^{1998}.91\)
\(S=91\left(3^0+3^6+...+3^{1998}\right)\)
\(S=\left(13.7\right)\left(3^0+3^6+...+3^{1998}\right)\)
\(S=13.7.\left(3^0+3^6+...+3^{1998}\right)\)(VÌ TÍCH CÓ THỪA SỐ 7 NÊN CHIA HẾT CHO 7 )
a) Nâng S thành 22S THÔI
b) nhóm 3 cặp lại
thế thôi
