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\(B=2+2^2+2^3+...+2^{2016}\)
\(2B=2^2+2^3+...+2^{2017}\)
\(B=2^{2017}-2\)
các ý khác tương tự
ý C nhân vs 3
D 4
E 5
3C = 3(1+3+3^2+.......+3^2017)
= 3+3^2+3^3+......+3^2018
3C - C = (3+3^2+3^3+......+3^2018) - (1+3+3^2+......+3^2017)
= 3^2018 - 1
=> C = (3^2018 - 1) : 2
còn lại tự làm nhé
$A=\dfrac12-\dfrac{2}{2^2}+\dfrac{3}{2^3}-\dfrac{4}{2^4}+\cdots+\dfrac{99}{2^{99}}-\dfrac{100}{2^{100}}$
Nhóm từng 2 số:
$A=\left(\dfrac12-\dfrac{2}{2^2}\right)+\left(\dfrac3{2^3}-\dfrac4{2^4}\right)+\cdots+\left(\dfrac{99}{2^{99}}-\dfrac{100}{2^{100}}\right)$
$=0+\dfrac18+\dfrac{2}{32}+\dfrac3{128}+\cdots+\dfrac{49}{2^{99}}$
$=\sum_{k=1}^{50}\dfrac{k-1}{2^{2k-1}}$
Ta có: $\dfrac{k-1}{2^{2k-1}}=\dfrac{2(k-1)}{4^k}$
Mà: $\sum_{k=1}^{\infty}\dfrac{k-1}{4^k}=\dfrac{1}{9}$
Nên: $A<2\cdot\dfrac19$ $=\dfrac29$
Vậy: $A<\dfrac29$
b)$4=1\cdot4,\quad28=4\cdot7,\quad70=7\cdot10,\ldots$
tức là: $E=\dfrac3{1\cdot4}+\dfrac3{4\cdot7}+\dfrac3{7\cdot10}+\cdots+\dfrac3{n(n+3)}$
Với $n=1,4,7,\ldots$.
Ta có: $\dfrac3{n(n+3)}=\dfrac1n-\dfrac1{n+3}$
Do đó: $E=\left(1-\dfrac14\right)+\left(\dfrac14-\dfrac17\right)+\left(\dfrac17-\dfrac1{10}\right)+\cdots+\left(\dfrac1n-\dfrac1{n+3}\right)$
$=1-\dfrac1{n+3}$
Vì: $\dfrac1{n+3}>0$ nên: $1-\dfrac1{n+3}<1$
Ta có:
$S=\dfrac{5}{1\cdot2\cdot3}+\dfrac{8}{2\cdot3\cdot4}+\dfrac{11}{3\cdot4\cdot5}+\cdots+\dfrac{6026}{2008\cdot2009\cdot2010}$
Nhận thấy tử số có dạng $3n+2$, nên:
$S=\sum_{n=1}^{2008}\dfrac{3n+2}{n(n+1)(n+2)}$
Ta có: $\dfrac{3n+2}{n(n+1)(n+2)}=\dfrac{1}{n(n+1)}+\dfrac{2}{(n+1)(n+2)}$
Do đó:
$S=\left(\dfrac1{1\cdot2}+\dfrac1{2\cdot3}+\cdots+\dfrac1{2008\cdot2009}\right)$
$+2\left(\dfrac1{2\cdot3}+\dfrac1{3\cdot4}+\cdots+\dfrac1{2009\cdot2010}\right)$
Mà: $\dfrac1{n(n+1)}=\dfrac1n-\dfrac1{n+1}$
Nên: $S=\left(1-\dfrac1{2009}\right)+2\left(\dfrac12-\dfrac1{2010}\right)$
$=1-\dfrac1{2009}+1-\dfrac1{1005}$
$=2-\dfrac1{2009}-\dfrac1{1005}$
Vì: $\dfrac1{2009}+\dfrac1{1005}>0$
Nên $S<2$
Câu 1:tính tổng
a) - 1 + 3 - 5 + 7 - . . . + 97 - 99
= - 1 + 3 + (-5) +7 + . . . + 97 + (-99)
=[ - 1 + 3 ] + [ - 5 + 7] + . . . + [-95 + 97] +-99
=2+2+... +2 +(-99)
=2.49 +(-99)
=98+(-99)
=(-1)
A= 1+2+22+23+24 / 1+2+22+23=20+....+24/ 20+....+23
=>rút gọn vế trái và vế phải ta được kết quả là ;24=16
B thì làm tương tự bài trên
chúc bạn họ tốt !
\(A=\frac{\left(1+2+2^2+2^3+2^4\right)}{\left(1+2+2^2+2^3\right)}\)
\(A=\frac{2^4\left(1+2+2^2+2^3\right)}{\left(1+2+2^2+2^3\right)}\)
\(A=2^4=16\)
\(B=\frac{\left(1+3+3^2+3^3+3^4\right)}{\left(1+3+3^2+3^3\right)}\)
\(B=\frac{3^4\left(1+3+3^2+3^3\right)}{\left(1+3+3^2+3^3\right)}\)
\(B=3^4=81\)
\(\text{Ta co }A=16< B=81\)
\(\Rightarrow A< B\)
Ta có:
A = \(\frac{1+2+2^2+2^3+2^4}{1+2+2^2+2^3}=\frac{\left(1+2+2^2+2^3\right)+2^4}{1+2+2^2+2^3}\)
= \(1+\frac{2^4}{1+2+2^2+2^3}=1+\frac{\left(1+2+2^2+2^3\right)+1}{1+2+2^2+2^3}\)
= \(1+1+\frac{1}{1+2+2^2+2^3}=2+\frac{1}{1+2+2^2+2^3}\)
B = \(\frac{1+3+3^2+3^3+3^4}{1+3+3^2+3^3}=\frac{\left(1+3+3^2+3^3\right)+3^4}{1+3+3^2+3^3}\)
= \(1+\frac{3^4}{1+3+3^2+3^3}=1+\frac{\left(1+3+3^2+3^3\right).2+1}{1+3+3^2+3^3}\)
= \(1+2+\frac{1}{1+3+3^2+3^3}=3+\frac{1}{1+3+3^2+3^3}\)
Ta thấy : 2 < 3 => \(2+\frac{1}{1+2+2^2+2^3}< 3+\frac{1}{1+3+3^2+3^3}\)
=> A < B