Có thể đề sẽ là \(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+\dfrac{1}{10.13}+...+\dfrac{1}{97.100}\\ =\dfrac{1}{3}\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+\dfrac{3}{10.13}+...+\dfrac{3}{97.100}\right)\\ =\dfrac{1}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\\ =\dfrac{1}{3}\left(1-\dfrac{1}{100}\right)=\dfrac{1}{3}.\dfrac{99}{100}=\dfrac{33}{100}\)

Bạn đánh lại chỗ 1/(j-4) nha.

KQ = 33/100

\(\frac{5}{4.7}+\frac{5}{7.10}+\frac{5}{10.13}+...+\frac{5}{97.100}\)

\(=5\left(\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+...+\frac{1}{97.100}\right)\)

\(=5.\frac{1}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{97}-\frac{1}{100}\right)\)

\(=\frac{5}{3}\left(\frac{1}{4}-\frac{1}{100}\right)\)

\(=\frac{2}{5}\)

thank you

\(\frac{6}{4\cdot7}+\frac{6}{7\cdot10}+\frac{6}{10\cdot13}+...+\frac{6}{97\cdot100}\)

\(=2\left(\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+\frac{3}{10\cdot13}+...+\frac{3}{97\cdot100}\right)\)

\(=2\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{97}-\frac{1}{100}\right)\)

\(=2\left(\frac{1}{4}-\frac{1}{100}\right)\)

\(=2\cdot\frac{24}{100}=\frac{24}{50}=\frac{12}{25}\)

Đặt \(A=\frac{6}{4\cdot7}+\frac{6}{7\cdot10}+\frac{6}{10\cdot13}+...+\frac{6}{97\cdot100}\)

\(A=\)\(2\cdot\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\right)\)

\(A=2\left(\frac{1}{4}-\frac{1}{100}\right)\)

\(A=2\left(\frac{25}{100}-\frac{1}{100}\right)\)

\(A=2\cdot\frac{24}{100}=2\cdot\frac{6}{25}=\frac{12}{25}\)

Đặt biểu thức là A.Từ biểu thức, ta suy ra:

\(A=6\cdot\left(\frac{1}{4\cdot7}+\frac{1}{7\cdot10}+\frac{1}{10\cdot13}+...+\frac{1}{97\cdot100}\right)\)

\(3A=6\cdot\left(\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+\frac{3}{10\cdot13}+...+\frac{3}{97\cdot100}\right)\)

\(3A=6\cdot\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{97}-\frac{1}{100}\right)\)

\(3A=6\cdot\left(\frac{1}{4}-\frac{1}{100}\right)\)

\(3A=6\cdot\frac{6}{25}\)

\(A=\frac{6}{25}:3=\frac{2}{25}\)

Vậy \(A=\frac{2}{25}\)

Là 3 phần 13 nhân 10 hay 3 phần 13 nhân 16 mới đúng ?

1+4-7+10-13+16-....-97+100-103

=1 + ( 4-7 ) + ( 10-13 ) + (16-19) + .... +( 94- 97) + (100 - 103)

= 1 -(3 + 3 +.....+ 3)

= 1 - 3*17 = -50

Bl:1+(4-7)+(10-13)+...+(100-103)

=1+(-2)+(-2)+...+(-2)

Có tất cả 34 số -2 nên:

=1+[(-2).34]

=1+(-68)

=-(68-1)

=-67(âm sáu bảy)

Theo đề ta có :

1+4-7+10-13+16-....-97+100-103

suy ra: 1+(4-7)+(10-13)+...+(100-103)=1+(-3)+(-3)+...+(-3). Có tất cả 34 số nên ta có: 1+(-3).34=1+(-112)=(-111).

-1 + 4 - 7 + 10 - 13 + 16 - ... - 97 + 100 - 103

= (-1 + 4) + (-7 + 10) + (-13 + 16) + ... + (-97 + 100) - 103

= 3 + 3 + 3 + ... + 3 - 103

= 51 - 103

= -52

S1=1-4+7-10+13-16+...+97-100

S1=(1+7+13+...+97)-(4+10+16+...+100)

S1=\(\frac{\left[\left(97-1\right):6+1\right].\left(97+1\right)}{2}-\frac{\left[\left(100-4\right):6+1\right].\left(100+4\right)}{2}\)

S1=\(\frac{17.98}{2}-\frac{17.104}{2}\)

S1=17.49-17.52

S1=17.(49-52)

S1=17.(-3)

S1=-51

      Vậy S1=-51

a)

Số số hạng là \(\left(101-1\right)\div1+1=101\) số hạng

Tổng là \(\left(101+1\right)\times101\div2=5151\) 

b)

Số số hạng là \(\left(100-7\right)\div3+1=32\) số hạng

Tổng là \(\left(100+7\right)\times32\div2=1712\)

\(1+2+3+4+5+...+101\)
\(=(101+1)+(100+2)+(99+3)+...\)
\(=(101+1)*\dfrac{(101-1):1+1}{2}\)
\(=102*50.5=5151\)
\(7+10+13+16+19+...+100\)
\(=(100+7)+(97+10)+(94+13)+...\)
\(=(100+7)*\dfrac{(100-7):3+1}{2}\)
\(=107*16=1712\)