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\(a.\) \(\frac{6^3+3.6^2+3^3}{-13}=\frac{2^3.3^3+3.3^2.2^2+3^3}{-13}=\frac{2^3.3^3+3^3.2^2+3^3}{-13}\)
\(=\frac{3^3.\left(2^3+2^2+1\right)}{-13}=\frac{3^3.13}{-13}=\frac{3^3.\left(-1\right)}{1}=-27\)
\(b.\)\(A=2^2+4^2+6^2+...+20^2=2^2\left(1+2^2+3^2+...+10^2\right)\)
\(A=2^2.\frac{10.\left(10+1\right).\left(2.10+1\right)}{6}=4.385=1540\)
( Ta có: công thức tính tổng bình phương liên tiếp tứ 1 đến n là: \(1^2+2^2+3^2+...+n^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6}\))
\(c.\)\(B=100^2+200^2+...+1000^2=\left(100.1\right)^2+\left(100.2\right)^2+...+\left(100.10\right)^2\)
\(B=100^2.1^2+100^2.2^2+...+100^2.10^2=100^2.\left(1^2+2^2+...+10^2\right)\)
Áp dụng công thức \(1^2+2^2+3^2+...+n^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6}\)
Ta có: \(B=100^2\times385=3,850,000\)
a) A = 2 + 23+25+...+249
=> 22.A = 23+25+27+...+251
22.A - A = 251-2
3A=251-2
\(A=\frac{2^{51}-2}{3}\)
b) B = 31-35+39-313+...-381
=> 34.B = 35 - 39+ 313 - 317+...-385
=> 34.B - B = -385-31
81B - B = -385-31
\(B=\frac{-3^{85}-3^1}{80}\)
c) C = -4-42-43-44-...-4100
=> 4C = -42-43-44-45-...-4101
=> 4C - C = -4101+4
3C = -4101+4
\(C=\frac{-4^{101}+4}{3}\)
a) \(D=\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...+\frac{1}{7^{100}}\)
\(\Rightarrow7D=1+\frac{1}{7}+\frac{1}{7^2}+...+\frac{1}{7^{99}}\)
\(\Rightarrow7D-D=\left(1+\frac{1}{7}+\frac{1}{7^2}+...+\frac{1}{7^{99}}\right)-\left(\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...+\frac{1}{7^{100}}\right)\)
\(\Rightarrow6D=1-\frac{1}{7^{100}}\)
\(\Rightarrow D=\left(1-\frac{1}{7^{100}}\right).\frac{1}{6}\)
( 7200+7199) : 7198 = 7200 : 7198 + 7199 : 7198 = 49 + 7 = 56
( 73+72)x(53+ 56)x ( 33x 3-92) = ( 73+72).(53+ 56) . ( 3^4 - 3^4) = ( 73+72).(53+ 56) . 0 = 0
câu c thì bn xem lại đề nha 23x7x53-( 52x 65+ 52x 35) hay 23x7x53-( 52x 65+ 52 x 35)
Dài quá . Làm 1 ý thôi nha .
\(A=1+2+2^2+...+2^{150}.\)
\(\Rightarrow2A=2+2^2+2^3+...+2^{151}\)
\(\Rightarrow2A-A=\left(2+2^2+.....+2^{151}\right)-\left(1+2+2^2+......+2^{150}\right)\)
\(\Rightarrow A=3^{151}-1\)
\(B=1+3+3^2+.....+3^{200}.\)
\(\Rightarrow3B=3+3^2+3^3+....+3^{201}\)
\(\Rightarrow3B-B=\left(3+3^2+.....+3^{201}\right)-\left(1+3+......+3^{200}\right)\)
\(\Rightarrow2B=3^{201}-1\)
\(\Rightarrow B=\frac{3^{201}-1}{2}\)
C=1+5+....+51000
=> 5C=5+52+......+51001