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\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}\)
\(\Rightarrow\dfrac{A}{3}=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\)
\(\Rightarrow A-\dfrac{A}{3}=\dfrac{2A}{3}=\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\dfrac{2A}{3}=\left(\dfrac{1}{3^2}-\dfrac{1}{3^2}\right)+\left(\dfrac{1}{3^3}-\dfrac{1}{3^3}\right)+...+\left(\dfrac{1}{3^{99}}-\dfrac{1}{3^{99}}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)=\dfrac{1}{3}-\dfrac{1}{3^{100}}\)
\(\Rightarrow2A=3\cdot\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\text{A}=\dfrac{1-\dfrac{1}{3^{99}}}{2}\)
\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}< \dfrac{1}{2}\)
a:
Số số hạng trong dãy M là:
(1002-12):10+1=100(số)
=>Sẽ có 50 cặp (1002;992); (982;972);....;(22;12) có hiệu bằng 10
\(M=1002-992+982-972+...+22-12\)
\(=\left(1002-992\right)+\left(982-972\right)+...+\left(22-12\right)\)
\(=10+10+...+10\)
=10*50=500
b: \(N=\left(202+182+...+42+22\right)-\left(192+172+...+32+12\right)\)
\(=\left(202-192\right)+\left(182-172\right)+...+\left(22-12\right)\)
=10+10+...+10
=10*10=100
Each term of S is n!(n2 + n + 1) = n![n(n + 1) + 1] = n(n + 1)n! + n!
By definition, n(n + 1)n! + n! = n! + n(n + 1)!
Therefore, S can be simplified as
1! + 1.2! + 2! + 2.3! + ... + 100! + 100.101!
So \(\dfrac{S+1}{101!}=\dfrac{1+1!+1\cdot2!+2!+2\cdot3!+...+100!+100\cdot101!}{101!}\)
\(=\dfrac{2!+1\cdot2!+2!+2\cdot3!+3!+...+100!+100\cdot101!}{101!}\)
\(=\dfrac{3!+2\cdot3!+3!+...+100!+100\cdot101!}{101!}\)
\(=\dfrac{4!+3\cdot4!+4!+...+100!+100\cdot101!}{101!}\)
\(=...\)
\(=\dfrac{100!+99\cdot100!+100!+100\cdot101!}{101!}\)
\(=\dfrac{101!+100\cdot101!}{101!}\)
\(=1+100=101\)
Hence, \(\dfrac{S+1}{101!}=101\)
Bài 1:
13 + 23 = 1 + 8 = 9 = 32 (là một số chính phương)
13 + 23 + 33 = 1 + 8 + 27 = 36 = 62 (là một số chính phương)
13 + 23 + 33 + 43 = 1 + 8 + 27 + 64 = 100 = 102 (là số cp)
13 + 23 + 33 + 43 + 53 = 1 + 8 + 27 + 64 + 125 = 225 = (15)2 là số cp
Bài 2:
1262 + 1 = \(\overline{..6}\) + 1 = \(\overline{...7}\) (không phải số chính phương)
100! + 8 = \(\overline{...0}\) + 8 = \(\overline{...8}\) (không phải là số chính phương)
1012 - 3 \(\overline{..01}\) - 3 = \(\overline{...8}\) (không phải là số chính phương)
107 + 7 = \(\overline{..0}\) + 7 = \(\overline{..7}\) (không phải là số chính phương)
11 + 112 + 113 = \(\overline{..1}\)+ \(\overline{..1}\)+ \(\overline{..1}\) = \(\overline{...3}\) (không phải số chính phương)
a) \(A=2+2^2+2^3+...+2^{2017}\)
\(2A=2^2+2^3+2^4+...+2^{2018}\)
\(2A-A=\left(2^2+2^3+2^4+...+2^{2018}\right)-\left(2+2^2+2^3+...+2^{2017}\right)\)
\(A=2^{2018}-2\)
b) \(C=1+3^2+3^4+...+3^{2018}\)
\(3^2\cdot C=3^2+3^4+3^6+...+3^{2020}\)
\(9C-C=\left(3^2+3^4+3^6+...+3^{2020}\right)-\left(1+3^2+3^4+...+3^{2018}\right)\)
\(8C=3^{2020}-1\)
\(\Rightarrow C=\dfrac{3^{2020}-1}{8}\)
\(Toru\)
\(B=1\cdot2\cdot3+2\cdot3\cdot4+...+\left(n-1\right)\cdot n\cdot\left(n+1\right)\)
=>\(4B=1\cdot2\cdot3\cdot4+2\cdot3\cdot4\cdot4+...+\left(n-1\right)\cdot n\left(n+1\right)\cdot4\)
=>\(4B=1\cdot2\cdot3\cdot4+2\cdot3\cdot4\left(5-1\right)+...+\left(n-1\right)\cdot n\left(n+1\right)\left[\left(n+2\right)-\left(n-2\right)\right]\)
=>\(4B=1\cdot2\cdot3\cdot4-1\cdot2\cdot3\cdot4+...+\left(n-2\right)\left(n-1\right)\cdot n\cdot\left(n+1\right)-\left(n-2\right)\cdot\left(n-1\right)\cdot n\cdot\left(n+1\right)+\left(n-1\right)\cdot n\left(n+1\right)\left(n+2\right)\)
=>\(4B=\left(n-1\right)\cdot n\cdot\left(n+1\right)\left(n+2\right)\)
=>\(B=\dfrac{\left(n-1\right)\cdot n\left(n+1\right)\left(n+2\right)}{4}\)
\(C=1\cdot4+2\cdot5+3\cdot6+...+n\left(n+3\right)\)
\(=1\cdot\left(1+3\right)+2\left(2+3\right)+...+n\left(n+3\right)\)
\(=\left(1^2+2^2+...+n^2\right)+3\left(1+2+...+n\right)\)
\(=\dfrac{n\left(n+1\right)\left(2n+1\right)}{6}+3\cdot\dfrac{n\left(n+1\right)}{2}\)
\(=\dfrac{n\left(n+1\right)\left(2n+1\right)}{6}+\dfrac{3n\left(n+1\right)}{2}\)
\(=\dfrac{n\left(n+1\right)}{2}\cdot\left(\dfrac{2n+1}{3}+3\right)\)
\(=\dfrac{n\left(n+1\right)}{2}\cdot\dfrac{2n+1+9}{3}\)
\(=\dfrac{n\left(n+1\right)\left(n+5\right)}{3}\)
\(D=1^2+2^2+...+n^2\)
\(=1+\left(1+1\right)\cdot2+\left(1+2\right)\cdot3+...+\left(1+n-1\right)\cdot n\)
\(=1+2+3+...+n+\left(1\cdot2+2\cdot3+...+\left(n-1\right)\cdot n\right)\)
Đặt \(A=1+2+3+...+n;E=1\cdot2+2\cdot3+...+\left(n-1\right)\cdot n\)
\(E=1\cdot2+2\cdot3+...+\left(n-1\right)\cdot n\)
=>\(3E=1\cdot2\cdot3+2\cdot3\cdot3+...+\left(n-1\right)\cdot n\cdot3\)
=>\(3E=1\cdot2\cdot3+2\cdot3\cdot\left(4-1\right)+...+\left(n-1\right)\cdot n\left[\left(n+1\right)-\left(n-2\right)\right]\)
=>\(3E=1\cdot2\cdot3-1\cdot2\cdot3+2\cdot3\cdot4+...+\left(n-1\right)\cdot n\left(n-2\right)-\left(n-1\right)\cdot n\left(n-2\right)+\left(n-1\right)\cdot n\cdot\left(n+1\right)\)
=>\(3E=\left(n-1\right)\cdot n\left(n+1\right)=n^3-n\)
=>\(E=\dfrac{n^3-n}{3}\)
\(A=1+2+3+...+n\)
Số số hạng là n-1+1=n(số)
Tổng của dãy số là: \(A=\dfrac{n\left(n+1\right)}{2}\)
=>\(D=\dfrac{n^3-n}{3}+\dfrac{n\left(n+1\right)}{2}\)
\(=\dfrac{2n^3-2n+3n^2+3n}{6}\)
=>\(D=\dfrac{2n^3+3n^2+n}{6}\)
