=1(2+1)+2(3+1)+3(4+1)+...+100(101+1)

=1.2+1+2.3+2+3.4+3+...+100.101+100

=(1.2+2.3+3.4+..+100.101)+(1+2+3+...+100)

=333300+5000

=338300

\(S=1.3+2.4+3.5+...+99.101\)

\(\Rightarrow S=1\left(2+1\right)+2\left(3+1\right)+...+99\left(100+1\right)\)

\(\Rightarrow S=\left(1.2+2.3+...+99.100\right)+\left(1+2+3+...+99\right)\)

Đặt \(A=1.2+2.3+...+99.100\)

\(\Rightarrow3A=1.2.3+2.3.\left(4-1\right)+...+99.100.\left(101-98\right)\)

\(\Rightarrow3S=1.2.3+2.3.4-1.2.3+...+99.100.101-98.99.100\)

\(\Rightarrow S=\frac{99.100.101}{3}\)

Đặt \(B=1+2+3+...+99\)

\(\Rightarrow B=\frac{\left(99+1\right)\left[\left(99-1\right):2+1\right]}{2}\)

\(\Rightarrow B=\frac{100.50}{2}=2500\)

\(\Rightarrow S=A+B=\frac{99.100.101}{3}+2500\)

S = 1 x 3 + 2 x 4 + 3 x 5 + ... + 99 x 101

S = ( 1 x 3 + 3 x 5 + ...+ 99 x 101) +  ( 2 x 4 + ...+ 98 x 100)

Đặt A = 1 x 3 + 3 x 5 + ...+ 99 x 101

=> 6 A = 1 x 3 x 6 + 3 x 5 x 6 + ...+ 99 x 101 x 6

6 A = 1 x 3 x ( 5+1) + 3 x 5 x ( 7-1) + ...+ 99 x 101 x ( 103 - 97)

6A = 1 x 3 x 5 + 1 x 3 + 3 x 5 x 7 - 1 x 3 x 5 + ...+ 99 x 101 x 103 - 97 x 99 x 101

6A = ( 1 x 3 + 1 x 3 x 5 + 3 x 5 x 7 +...+ 99 x 101 x 103) - ( 1 x 3 x 5 + ...+ 97 x 99 x 101)

6A = 1 x  3 + 99 x 101 x 103

\(\Rightarrow A=\frac{1.3+99.101.103}{6}=171650\)

Đặt B = 2 x 4 + ...+ 98 x 100

=> 6B = 2 x 4 x 6 + 4 x 6 x 6 + ...+ 98 x 100 x 6

6B = 2 x 4 x 6 + 4 x 6 x ( 8-2) + ...+ 98 x 100 x ( 102 - 96)

6B = 2 x 4 x 6 + 4 x6 x8 - 2x4x6 + ...+ 98x100x102 - 96x98x100

6B = ( 2 x 4 x 6 + 4 x 6 x 8 +...+98x100x102) - ( 2x4x6+...+96x98x100)

6B = 98 x 100 x 102

\(\Rightarrow B=\frac{98.100.102}{6}=166600\)

Thay A;B vào S, có
S = 171 650 + 166 600

S = 338 250

= 338250

Học tốt

khó dữ vậy ba ?????

bằng 333300

Lời giải:
Xét thừa số tổng quát $1+\frac{1}{n(n+2)}=\frac{n(n+2)+1}{n(n+2)}=\frac{(n+1)^2}{n(n+2)}$

Khi đó:

$1+\frac{1}{1.3}=\frac{2^2}{1.3}$

$1+\frac{1}{2.4}=\frac{3^2}{2.4}$

.........

$1+\frac{1}{99.101}=\frac{100^2}{99.101}$

Khi đó:

$A=\frac{2^2.3^2.4^2......100^2}{(1.3).(2.4).(3.5)....(99.101)}$

$=\frac{(2.3.4...100)(2.3.4...100)}{(1.2.3...99)(3.4.5...101)}$

$=\frac{2.3.4...100}{1.2.3..99}.\frac{2.3.4...100}{3.4.5..101}$
$=100.\frac{2}{101}=\frac{200}{101}$

giúp em với

\(\left(1+\frac{1}{1\times3}\right)\times\left(1+\frac{1}{2\times4}\right)\times\left(1+\frac{1}{3\times5}\right)\times...\times\left(1+\frac{1}{99.101}\right)\)

\(=\left(\frac{3}{3}+\frac{1}{3}\right)\times\left(\frac{8}{8}+\frac{1}{8}\right)\times\left(\frac{15}{15}+\frac{1}{15}\right)\times...\times\left(\frac{9999}{9999}+\frac{1}{9999}\right)\)

\(=\frac{4}{3}\times\frac{9}{8}\times\frac{16}{15}\times...\times\frac{10000}{9999}\)

\(=\frac{4\times9\times16\times...\times10000}{3\times8\times15\times...\times9999}\)

\(=\frac{2\times2\times3\times3\times4\times4\times...\times100\times100}{1\times3\times2\times4\times3\times5\times...\times99\times101}\)

\(=\frac{2\times100}{101}=\frac{200}{101}\)

mk cx co dap an vay

$\dfrac{1}{2+3}$