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#)Giải :
\(A=1+2+2^2+...+2^{100}\)
\(2A=2+2^2+2^3+...+2^{101}\)
\(2A-A=\left(2+2^2+2^3+...+2^{101}\right)-\left(1+2+2^2+...+2^{100}\right)\)
\(A=2^{101}-1\)
\(B=1+3^2+3^4+...+3^{100}\)
\(3^2B=3^2+3^4+3^6+...+3^{102}\)
\(3^2B-B=\left(3^2+3^4+3^6+...+3^{102}\right)-\left(1+3^2+3^4+...+3^{100}\right)\)
\(8B=3^{102}-1\)
\(B=\frac{3^{102}-1}{8}\)
\(C=1+5^3+5^6+...+5^{99}\)
\(5^2C=5^3+5^6+5^9+...+5^{102}\)
\(5^2C-C=\left(5^3+5^6+5^9...+5^{102}\right)-\left(1+5^3+5^6+...+5^{99}\right)\)
\(24C=5^{102}-1\)
\(C=\frac{5^{102}-1}{24}\)
a) A = 1 + 22 + ... + 2100
=> 2A = 22 + 23 + ... + 2101
Lấy 2A - A = (2 + 22 + ... + 2101) - (1 + 22 + ... 2100)
A = 2101 - 1
b) B = 1 + 32 + 34 + ... + 3100
=> 32B = 32 + 34 + 36 + ..... + 3102
=> 9B = 32 + 34 + 36 + ..... + 3102
Lấy 9B - B = ( 32 + 34 + 36 + ..... + 3102) - (1 + 32 + 34 + ... + 3100)
8B = 3102 - 1
B = \(\frac{3^{102}-1}{8}\)
c) C = 1 + 53 + 56 + ... + 599
=> 53.C = 53 . 56 . 59 + ... + 5102
=> 125.C = 53 . 56 . 59 + ... + 5102
Lấy 125.C - C = (53 . 56 . 59 + ... + 5102) - (1 + 53 + 56 + ... + 599)
124.C = 5102 - 1
=> C = \(\frac{5^{102}-1}{124}\)
theo mình nghĩ là như th61 này
\(2\cdot2^{99}-2^{99}=2^{99}\)
\(2^{99}=2\cdot2^{98}\)
\(2\cdot2^{98}-2^{98}=2^{98}\)
vậy tức là \(2^n-2^{n-1}=2^{n-1}\)
đến cuối bạn sẽ có \(2^3-2^2=4\)
4-2-1=1
Ta có:
$A=2^{100}-2^{99}-2^{98}-\cdots-2^2-2-1$
$=2^{100}-(2^{99}+2^{98}+\cdots+2+1)$
Mà: $2^{99}+2^{98}+\cdots+2+1=2^{100}-1$
Nên: $A=2^{100}-(2^{100}-1)$ $=1$
\(\frac{101+100+99+98+...+3+2+1}{101-100+99-98+...+3-2+1}\)
\(=\frac{\frac{101.102}{2}}{51}\)
\(=101\)
$A=\dfrac12-\dfrac{2}{2^2}+\dfrac{3}{2^3}-\dfrac{4}{2^4}+\cdots+\dfrac{99}{2^{99}}-\dfrac{100}{2^{100}}$
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: $1-\dfrac1{n+3}<1$
1) [1+(-2)]+[3+(-4)]+...+ [19+(-20)]
=(-1)+ (-1)+ ...+ (-1)
= -10
2) (1-2)+(3-4)+... +(99-100)
= (-1)+ (-1)+... + (-1)
= -50
1) =[1+(-2)]+[3+(-4)]+....+[19+9-20)]=-1+(-1)+....+(-1)=10 2) =(1-2)+(3-4)+....+(99-100)=-1+(-1)+...+(-1)=50