Tính:
a ) A = 1 1.2 + 1 2.3 + 1 3.4 + ... + 1 2017.2018 b ) B = 3 1.3 + 3 3.5 + 3 5.7 + ... + 3 199.201
Tính tổng: M=1.2+2.3+....+48.49 N=1+2+...+48 A=1.2+2.3+...+99.100 Cảm ơn
b: Tổng của N là:
\(\dfrac{49\cdot48}{2}=49\cdot24=1176\)
a) \(3M=1.2.3+2.3.3+...+48.49.3=1.2.3+2.3.\left(4-1\right)+...+48.49.\left(50-47\right)=1.2.3+2.3.4-1.2.3+...+48.49.50-47.48.49=48.49.50\Rightarrow M=\dfrac{48.49.50}{3}\Rightarrow M=39200\)
b) Tương tự câu a
cho A = 1/1.2+1/2.3+1/3.4+...+1/49.50 ; cho B = 1.2+1.3+3.4+....+49.50
tính 50mủ 2A - B/17
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{50-49}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}=\frac{49}{50}\)
\(B=1.2+2.3+3.4+...+49.50\)
\(3B=1.2.3+2.3.3+3.4.3+...+49.50.3\)
\(=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+49.50.\left(51-48\right)\)
\(=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+49.50.51-48.49.50\)
\(=49.50.51\)
\(B=\frac{49.50.51}{3}=49.50.17\)
\(50^2.A-\frac{B}{17}=49.50-49.50=0\)
1.Tính tổng: A = 1.2 + 3.4 +...+ 2(2n+1)(n+1)
2.Tính tổng: A = 1.3 + 3.7 + 5.11 +...+ 99.199
tính tổng A=1/1.2+1/2.3+..........+1/2003.2004
=1-1/2+1/2-1/3+...+1/2003-1/2004
=1-1/2004
=2003/2004
Lời giải:
$A=\frac{2-1}{1.2}+\frac{3-2}{2.3}+...+\frac{2004-2003}{2003.2004}$
$=\frac{2}{1.2}-\frac{1}{1.2}+\frac{3}{2.3}-\frac{2}{2.3}+....+\frac{2004}{2003.2004}-\frac{2003}{2003.2004}$
$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-....+\frac{1}{2003}-\frac{1}{2004}$
$=1-\frac{1}{2004}=\frac{2003}{2004}$
tính A = 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/2013.2014
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2013.2014}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2013}-\dfrac{1}{2014}\\ =1-\dfrac{1}{2014}\\ =\dfrac{2013}{2014}\)
Tính: A= 1 - 1/1.2 - 1/2.3 - 1/3.4 - ...- 1/97.98
A = 1 - \(\dfrac{1}{1.2}\) - \(\dfrac{1}{2.3}-\dfrac{1}{3.4}-\dfrac{1}{4.5}...-\dfrac{1}{97.98}\)
A= 1-\(\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{97.98}\right)\)
A= 1- \(\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}...+\dfrac{1}{97}-\dfrac{1}{98}\right)\)
A= 1- \(\left(\dfrac{1}{1}-\dfrac{1}{98}\right)\)
A=1- 1 + \(\dfrac{1}{98}\)
A= \(\dfrac{1}{98}\)
Lời giải:
$1-A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{97.98}$
$1-A=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{98-97}{97.98}$
$1-A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{97}-\frac{1}{98}$
$=1-\frac{1}{98}$
$\Rightarrow A=\frac{1}{98}$
Tính tổng: A = 1.2 + 3.4 +...+ 2(2n+1)(n+1)
Bài 1: Tính A = 1.2 + 2.3 + 3.4 +...+n. (n+1)
=> Ta thấy rằng mỗi số hạng trong dãu số trên đều là tích của hai số tự nhiên liên tiếp, khi đó:
Gọi a1 = 1.2 → 3a1 = 1.2.3 → 3a1 = 1.2.3 - 0.1.2
Tương tự:
a2 = 2.3 → 3a2 = 2.3.3 → 3a2 = 2.3.4 - 1.2.3
a3 = 3.4 → 3a3 = 3.3.4 → 3a3 = 3.4.5 - 2.3.4 ....
a(n - 1) = (n - 1).n → 3a(n - 1) = 3(n - 1)n → 3a(n - 1) = (n - 1).n.(n + 1) - (n - 2).(n - 1).n
an = n.(n - 1) → 3an = 3n(n + 1) → 3an = n(n + 1)(n + 2) - (n - 1)n(n + 1)
Cộng vế với vế của các đẳng thức trên ta được:
3(a1 + a2 + a3 +...+ an) = n(n + 1)(n + 2)
-> A = n.(n+1) .( n+2) / 3
Lời giải:
$A=1.2+2.3+3.4+...+n(n+1)$
$3A=1.2.3+2.3.3+3.4.3+....+n(n+1).3$
$3A=1.2.3+2.3(4-1)+3.4(5-2)+....+n(n+1)[(n+2)-(n-1)]$
$3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+n(n+1)(n+2)-(n-1)n(n+1)$
$=[1.2.3+2.3.4+3.4.5+....+n(n+1)(n+2)]-[1.2.3+2.3.4+....+(n-1)n(n+1)]$
$=n(n+1)(n+2)$
$\Rightarrow A=\frac{n(n+1)(n+2)}{3}$
Tính A = 1.2 + 2.3 + 3.4 + … + n.(n + 1)
Tham khảo:
https://olm.vn/hoi-dap/detail/7327860996.html
Ta có:
\(3A=1.2.3+2.3.3+3.4.3+....+n\left(n+1\right).3\)
\(\Leftrightarrow3A=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+n\left(n+1\right)\left[\left(n+2\right)-\left(n-1\right)\right]\)
\(\Leftrightarrow3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+n\left(n+1\right)\left(n+2\right)-\left(n-1\right)n\left(n+1\right)\)
\(\Leftrightarrow3A=n\left(n+1\right)\left(n+2\right)\)
\(\Rightarrow A=\dfrac{n\left(n+1\right)\left(n+2\right)}{3}\)
tính tổng: A= 1.2 + 2.3 + ... + n.(n+1)
A = 1.2 + 2.3 +...+ n.(n+1)
1.2.3 = 1.2.3
2.3.3 = 2.3.( 4-1) = 2.3.4 - 1.2.3
3.4.3 = 3.4(5-2) = 3.4.5 - 2.3.4
.................................................
n(n+1).3 =n(n+1)[ (n+2) - (n-1)] = n(n+1)(n+2) - (n-1)n(n+1)
Cộng vế với vế ta có:
1.2.3+2.3.3+...+n(n+1).3 = n(n+1)(n+2)
3.[1.2+ 2.3+...+ n(n+1)] = n(n+1)(n+2)
1.2 + 2.3 +...+n(n+1) = n(n+1)(n+2): 3
A = 1.2 + 2.3 +...+ n.(n+1)
1.2.3 = 1.2.3
2.3.3 = 2.3.( 4-1) = 2.3.4 - 1.2.3
3.4.3 = 3.4(5-2) = 3.4.5 - 2.3.4
.................................................
n(n+1).3 =n(n+1)[ (n+2) - (n-1)] = n(n+1)(n+2) - (n-1)n(n+1)
Cộng vế với vế ta có:
1.2.3+2.3.3+...+n(n+1).3 = n(n+1)(n+2)
3.[1.2+ 2.3+...+ n(n+1)] = n(n+1)(n+2)
1.2 + 2.3 +...+n(n+1) = n(n+1)(n+2): 3
HT!