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\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{49\cdot50}\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(A=\frac{1}{1}-\frac{1}{50}\)
\(A=\frac{49}{50}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}=\frac{49}{50}\)
Bài giải:
3S = 1.2.3 + 2.3.3 + 3.4.3 + ... + n(n +1)3
= 1.2.(3 - 0) + 2.3.(4 - 1) + 3.4.(5 - 2) + ...+ n(n + 1)[(n + 2) - (n -1)]
= 1.2.3 + 2.3.4 - 2.3 + 3.4.5 - 2.3.4 + ... + n(n + 1)(n + 2) - n(n + 1)(n - 1)
= n(n + 1)(n + 2)
=> S N(N+1)(n+2)/3
3S = 1.2.3 + 2.3.3 + 3.4.3 + ... + n(n +1)3
= 1.2.(3 - 0) + 2.3.(4 - 1) + 3.4.(5 - 2) + ...+ n(n + 1)[(n + 2) - (n -1)]
= 1.2.3 + 2.3.4 - 2.3 + 3.4.5 - 2.3.4 + ... + n(n + 1)(n + 2) - n(n + 1)(n - 1)
= n(n + 1)(n + 2)
=> S = \(\frac{n\left(n+1\right)\left(n+2\right)}{3}\)
\(=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{n+1-n}{n.\left(n+1\right)}.\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n}-\frac{1}{n+1}=1-\frac{1}{n+1}< 1\)
D = 1.2 + 2.3+ 3.4 +...+ 99.100
=>3D=1.2.3+2.3.3+3.4.3+...+99.100.3
=1.2.(3-0)+2.3.(4-1)+3.4.(5-2)+....+99.100.(101-98)
=1.2.3-0.1.2+2.3.4-1.2.3+3.4.5-2.3.4+...+99.100.101-98.99.100
=99.100.101-0.1.2
=99.100.101
=999900
=>D=999900:3=333300
Dn = 1.2 + 2.3 + 3.4 +...+ n (n +1)
=>3Dn=1.2.3+2.3.3+3.4.3+...+n(n+1).3
=1.2.(3-0)+2.3.(4-1)+3.4.(5-2)+...+n.(n+1).[(n+2)-(n-1)]
=1.2.3-0.1.2+2.3.4-1.2.3+2.3.4-2.3.4+....+n(n+1)(n+2)-(n-1)n(n+1)
=n.(n+1).(n+2)-0.1.2
=n.(n+1)(n+2)
=>Dn=n.(n+1)(n+2):3
=>điều cần chứng minh
a) Đặt A = 1.2 + 2.3 + ........ + (n-1)n
3A = 1.2.3 + 2.3.(4-1) + .... + (n-1)n[(n+1)-(n-2)]
3A = 1.2.3 + 2.3.4 - 1.2.3 + .... + (n-1)n(n+1) - (n-2)(n-1)n
3A = (1.2.3 - 1.2..3) + ... + (n-1)n(n+1)
A = \(\frac{\left(n-1\right)n\left(n+1\right)}{3}\)
b) Đặt B = 12 + 22 + ..... + n2
B = 1(2 - 1) + 2(3 - 1) + ..... + n[(n + 1) - 1]
B = 1.2 + 2.3 + .......... + n(n + 1) - (1+2+3+....+n)
B = A - \(\frac{n\left(n+1\right)}{2}\)
Lời giải:
$A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{n(n+1)}$
$=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+....+\frac{(n+1)-n}{n(n+1)}$
$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n}-\frac{1}{n+1}$
$=1-\frac{1}{n+1}=\frac{n}{n+1}$
Ta có đpcm.
\(A=1.2+2.3+3.4+.......+\left(n-1\right).n\)
\(\Rightarrow3A=1.2.3+2.3.3+3.4.3+......+\left(n-1\right).n.3\)
\(=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+.....+\left(n-1\right).n.\left[\left(n+1\right)-\left(n-2\right)\right]\)
\(=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+......+\left(n-1\right).n\left(n+1\right)-\left(n-1\right).n\left(n-2\right)\)
\(=\left(n-1\right).n.\left(n+1\right)\)
\(\Rightarrow A=\frac{\left(n-1\right).n.\left(n+1\right)}{3}\)( đpcm )
1/A = 1 + 2 + 3 + 4 +.......+ n
Hay A = n + ... + 4 + 3 + 2 + 1 (Viết ngược lại )
=> A + A = (1 + n) + ... + (n + 1) Có n cặp
=> 2.A = (1 + n).n
=> A = (1 + n).n/2 => Đpcm
2/ B=1.2+2.3+3.4.....+(n-1).n
ta có
3.B=1.2.(3-0)+2.3.(4-1)+3.4.(5 -2)...+ (n-1).n . ((n+1) - (n-2))
3.B=1.2.3+2.3.4+3.4.5+...+ (n-1) . n. (n+1) - 0.1.2 -1.2.3 -2.3.4 -3.4.5 -...(n-1)(n+1) n
3A=n.(n-1).(n+1)
A=1/3.n.(n-1).(n+1)
a) \(VP=\frac{1}{n}-\frac{1}{n+1}=\frac{n+1}{n\left(n+1\right)}-\frac{n}{n\left(n+1\right)}=\frac{n+1-n}{n\left(n+1\right)}=\frac{1}{n\left(n+1\right)}\)
VT=VP=>đpcm
b)áp dụng a)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+..+\frac{1}{99}-\frac{1}{100}=\frac{1}{1}-\frac{1}{100}=\frac{100}{100}-\frac{1}{100}=\frac{99}{100}\)
Vậy A=99/100
b) A=1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100
=1-1/100
=99/100
=9,9