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Đặt A=1/10+1/40+1/88+1/154+1/238+1/340
A=1/2.5+1/5.8+1/8.11+1/11.14+1/14.17+1/17.20
3A=3/2.5+3/5.8+....+3/17.20
3A=1/2-1/5+1/5-1/8+...+1/17-1/20
3A=1/2-1/20
3A=9/20
2)
Giữ nguyên p/s 1/2^2
Ta có:1/3^2<1/2.3
1/4^2<1/3.4
...............
1/n^2<1/(n-1).n
=>1/3^2+1/4^2+...+1/n^2<1/2.3+1/3.4+...+1/(n-1).n
=>1/3^2+1/4^2+.....+1/n^2<1/2-1/3+1/3-1/4+.........+1/n-1-1/n
=>1/2^2+1/3^2+.....+1/n^2<1/2^2+1/2-1/n
=>1/2^2+1/3^2+....+1/n^2<3/4-1/n<3/4
3)
2B=2/3.5+2/5.7+....+2/47.49+2/49.51
2B=1/3-1/5+1/5-1/7+.....+1/47-1/49+1/49-1/51
2B=1/3-1/51
2B=16/51
B=16/51:2
B=8/51
A=1+1/2+1/2^2+...+1/2^2010
2A=2+1+1/2+....+1/2^2009
2A-A=(2+1+1/2+...+1/2^2009)-(1+1/2+1/2^2+....+1/2^2010)
A=2-1/2^2010
\(a,S=1+3+3^2+....+3^{100}.\)
\(\Rightarrow3S=3+3^2+...+3^{101}\)
\(\Rightarrow3S-S=\left(3+3^2+...+3^{101}\right)-\left(1+3+....+3^{100}\right)\)
\(\Rightarrow2S=3^{101}-1\)
\(\Rightarrow S=\frac{3^{101}-1}{2}\)
\(b,A=1+3^2+3^4+...+3^{100}\)
\(\Rightarrow3^2A=3^2+3^4+...+3^{102}\)
\(\Rightarrow9A-A=\left(3^2+3^4+...+3^{102}\right)-\left(1+3^2+....+3^{100}\right)\)
\(\Rightarrow8A=3^{102}-1\)
\(\Rightarrow A=\frac{3^{102}-1}{8}\)
a) 3A = 3 + 3^2 + 3^3 + 3^4 + ... + 3^100 + 3^ 101
=> 3A - A = (3 + 3^2 + 3^3 + 3^4 + ... + 3^100 + 3^ 101) - (1 + 3 + 3 ^2 + 3 ^ 3 + ... + 3 ^100)
=> 2A = 3^101 - 1 => A = (3^101 - 1)/2
b) 4B = 4 + 4 ^ 2 + 4 ^3 + 4 ^ 4 + ... + 4 ^ 100 + 4^ 101
=> 4B - B = (4 + 4 ^ 2 + 4 ^3 + 4 ^ 4 + ... + 4 ^ 100 + 4^ 101) - (1 + 4 + 4 ^ 2 + 4 ^3 + 4 ^ 4 + ... + 4 ^ 100 )
=> 3B = 4^101 - 1 => B = ( 4^101 - 1)/2
c) xem lại đề ý c xem quy luật như thế nào nhé.
d) 3D = 3^101 + 3^ 102 + 3^ 103 + ... + 36 150 + 3^ 151
=> 3D - D = (3^101 + 3^ 102 + 3^ 103 + ... + 36 150 + 3^ 151) - (3 ^100 + 3 ^ 101 + 3 ^ 102 + .... + 3 ^ 150)
=> 2D = 3^ 151 - 3^100 => D = ( 3^ 151 - 3^100)/2
a) Có A=\(1+3+3^2+3^3+....+3^{100}\)
\(\Rightarrow\)3A =\(3\left(1+3+3^2+3^3+...+3^{100}\right)\)=\(3+3^2+3^3+3^4+...+3^{101}\)
\(\Rightarrow2A=3+3^2+3^3+....+3^{101}-1-3-3^2-3^3-....-3^{100}=3^{101}-1\)\(\Rightarrow A=\dfrac{3^{101}-1}{2}\)
Bài b/c/d : bn cứ lm tương tự.