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\(a,\)Đặt \(A=1+2+2^2+...+2^{99}+2^{100}\)
\(\Rightarrow2A=2+2^2+...+2^{100}+2^{101}\)
\(\Rightarrow2A-A=\left(2+2^2+2^3+...+2^{101}\right)-\left(1+2+2^2+...2^{100}\right)\)
\(\Rightarrow A=2^{101}-1\)
\(b,\)Đặt \(B=5+5^3+5^5+...+5^{97}+5^{99}\)
\(\Rightarrow5^2B=5^3+5^5+...+5^{99}+5^{101}\)
\(\Rightarrow25B-B=\left(5^3+5^5+...+5^{99}+5^{101}\right)-\left(5+5^3+...+5^{99}\right)\)
\(\Rightarrow24B=5^{101}-5\)
\(\Rightarrow B=\frac{5^{101}-5}{24}\)
Ta có:A=\(1+3+3^2+3^3+...+3^{2012}\)
3A=\(3\cdot\left(1+3+3^2+3^3+...+3^{2012}\right)\)
3A=\(3+3^2+3^3+3^4+...+3^{2013}\)
3A-A=\(\left(3+3^2+3^3+3^4+...+3^{2013}\right)-\left(1+3+3^2+3^3+...+3^{2012}\right)\)
2A=\(3+3^2+3^3+3^4+...+3^{2013}-1-3-3^2-3^3-...-3^{2012}\)
2A=\(\left(3-3\right)+\left(3^2-3^2\right)+\left(3^3-3^3\right)+...+\left(3^{2012}-3^{2012}\right)+\left(3^{2013}-1\right)\)
2A=\(0+0+0+...+0+3^{2013}-1\)
2A=\(3^{2013}-1\)
A=\(\frac{3^{2013}-1}{2}\)
B=\(3^{2013}\div2\)
B=\(\frac{3^{2013}}{2}\)
VậyB-A=\(\frac{3^{2013}}{2}-\frac{3^{2013}-1}{2}\)
\(B-A=\frac{3^{2013}-\left(3^{2013}-1\right)}{2}\)
\(B-A=\frac{3^{2013}-3^{2013}+1}{2}\)
\(B-A=\frac{1}{2}=0,5\)
a/ta có:s=(1-3+32-33)+.................+(396-397+398-399)
=-20+.....................+396.(-20.(1+...................396))
suy ra s chia het cho -20
b/ 3s=3-32+33-34+.................+399-3100
3s+s=(3-32+33-34+..........................+399-3100 +(1-3+32-33)+............+398-399)
4s=1-3100
s=(1-3100):4
vì s chia hết cho -20 suy ra s chia hết cho 4 suy ra 1-3100 chia hêt cho 4 suy ra 3100:4 dư 1
nếu đúng thì tíc cho mình 2 cái nhé!
\(A=3+3^2+3^3+...+3^{100}\)
\(3A=3^2+3^3+3^4+...+3^{101}\)
\(3A-A=\left(3^2+3^3+...+3^{101}\right)-\left(3+3^2+...+3^{100}\right)\)
\(2A=3^{101}-3\)
\(A=\left(3^{101}-3\right):2\)
Ta có : \(2A+3=3^{101}\)
\(→n=101\)
~ Ủng hộ nhé ~
a) \(D=\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...+\frac{1}{7^{100}}\)
\(\Rightarrow7D=1+\frac{1}{7}+\frac{1}{7^2}+...+\frac{1}{7^{99}}\)
\(\Rightarrow7D-D=\left(1+\frac{1}{7}+\frac{1}{7^2}+...+\frac{1}{7^{99}}\right)-\left(\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...+\frac{1}{7^{100}}\right)\)
\(\Rightarrow6D=1-\frac{1}{7^{100}}\)
\(\Rightarrow D=\left(1-\frac{1}{7^{100}}\right).\frac{1}{6}\)
A = 1 + 3 + 32 + 33 + ... + 399
3A = 3 + 32 + 33 + .. + 3100
3A -A = 3 + 32 + 33 + ... + 3100 - 1 - 3 - 32 - 399
2A = 3100 - 1
B - 2A = 3100 - ( 3100 - 1 ) = 1