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Bạn lấy số cuối dãy trừ số đầu dãychia khoảng cách la 1 va tru 1 la tim ra so phan tu
Roi ban lay so dau day cong so cuoi daycong phan tu ban vua tinh ra chia 2


A=7.(1/10.11+1/11.12+...+1/69.70)
A=7.(1/10-1/11+1/11-1/12+...+1/69-1/70)
A=7.(1/10-1/70)
A=7. 3/35
A= 3/5
chúc bạn học tốt nha
\(A=\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)
\(\Rightarrow A=7\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{69}-\frac{1}{70}\right)\)
\(\Rightarrow A=7\left(\frac{1}{10}-\frac{1}{70}\right)\)
\(\Rightarrow A=7.\frac{6}{70}=\frac{6}{10}=\frac{3}{5}\)

\(A=10\cdot11+11\cdot12+12\cdot13+...+29\cdot30\\ 3A=10\cdot11\cdot3+11\cdot12\cdot3+12\cdot13\cdot3+...+29\cdot30\cdot3\\ 3A=10\cdot11\cdot\left(12-9\right)+11\cdot12\cdot\left(13-10\right)+12\cdot13\cdot\left(14-11\right)+...+29\cdot30\cdot\left(31-28\right)\\ 3A=10\cdot11\cdot12-9\cdot10\cdot11+11\cdot12\cdot13-10\cdot11\cdot12+12\cdot13\cdot14-11\cdot12\cdot13+...+29\cdot30\cdot31-28\cdot29\cdot30\\ 3A=\left(10\cdot11\cdot12-10\cdot11\cdot12\right)+\left(11\cdot12\cdot13-11\cdot12\cdot13\right)+...+\left(28\cdot29\cdot30-28\cdot29\cdot30\right)+29\cdot30\cdot31-9\cdot10\cdot11\\ 3A=29\cdot30\cdot31-9\cdot10\cdot11\\ A=\dfrac{29\cdot30\cdot31-9\cdot10\cdot11}{3}\\ A=\dfrac{3\cdot\left(29\cdot10\cdot31-3\cdot10\cdot11\right)}{3}\\ A=29\cdot10\cdot31-3\cdot10\cdot11\\ A=10\cdot\left(29\cdot31-3\cdot11\right)\\ A=10\cdot866\\ A=8660\)

Lời giải:
$3S=10.11(12-9)+11.12(13-10)+12.13(14-11)+...+98.99(100-97)+99.100(101-98)$
$=(10.11.12+11.12.13+12.13.14+...+98.99.100+99.100.101)-(9.10.11+10.11.12+...+97.98.99+98.99.100)$
$=99.100.101-9.10.11$
$\Rightarrow S=\frac{99.100.101-9.10.11}{3}=33.100.101-3.10.11$

\(\dfrac{3}{10.11}\) + \(\dfrac{3}{11.12}\) + \(\dfrac{3}{12.13}\) + \(\dfrac{3}{13.14}\) + \(\dfrac{3}{14.15}\)
= \(\dfrac{3}{10}\) - \(\dfrac{3}{11}\) + \(\dfrac{3}{11}\) - \(\dfrac{3}{12}\) + \(\dfrac{3}{12}\) - \(\dfrac{3}{13}\) + \(\dfrac{3}{13}\) - \(\dfrac{3}{14}\) + \(\dfrac{3}{14}\) - \(\dfrac{3}{15}\)
= \(\dfrac{3}{10}\) - \(\dfrac{3}{15}\) = \(\dfrac{1}{10}\)
\(\dfrac{3}{10.11}+\dfrac{3}{11.12}+\dfrac{3}{12.13}+\dfrac{3}{13.14}+\dfrac{3}{14.15}\)
\(=\dfrac{3}{1}.\left(\dfrac{3}{10}-\dfrac{3}{11}+\dfrac{3}{11}-\dfrac{3}{12}+\dfrac{3}{12}-\dfrac{3}{13}+\dfrac{3}{13}-\dfrac{3}{14}+\dfrac{3}{14}-\dfrac{3}{15}\right)\)
\(=\dfrac{3}{1}.\left(\dfrac{3}{10}-\dfrac{3}{15}\right)\)
\(=\dfrac{3}{1}.\left(\dfrac{9}{30}-\dfrac{6}{30}\right)\)
\(=\dfrac{3}{1}.\dfrac{1}{10}\)
\(=\dfrac{3}{10}\)

tổng này có:99-10+1=90 so hang
(10,10+99,99)x90:2=4954,05
k nhe

10.11+11.12+12.13+...+97.98+98.99+99.100
=10-11+11-12+12-13+...+97-98+98-99+99-100
=10-100
=-90
Đặt A = 10.11 + 11.12 + ... + 98.99 + 99.100
3A = 10.11.3 + 11.12.3 + ... + 98.99.3 + 99.100.3
3A = 10.11.(12 -9) + 11.12.(13-10) + ... + 98.99.(100 - 97) + 99.100.(101-98)
3A = 10.11.12 - 9.10.11 + 11.12.13 - 10.11.12 + ... + 98.99.100 - 97.98.99 + 99.100.101 - 98.99.100
3A = (10.11.12 + 11.12.13 + ... + 98.99.100 + 99.100.101) - (9.10.11 + 10.11.12 + ... + 97.98.99 + 98.99.100)
3A = 99.100.101 - 9.10.11
3A = 999799
A = 999799 : 3
\(A=10\cdot11+11\cdot12+12\cdot13+...+499\cdot500\\ 3A=10\cdot11\cdot\left(12-9\right)+11\cdot12\cdot\left(13-10\right)+12\cdot13\cdot\left(14-11\right)+...+499\cdot500\cdot\left(501-498\right)\\ 3A=10\cdot11\cdot12-9\cdot10\cdot11+11\cdot12\cdot13-10\cdot11\cdot12+12\cdot13\cdot14-11\cdot12\cdot13+...+499\cdot500\cdot501-498\cdot499\cdot500\\ 3A=\left(10\cdot11\cdot12-10\cdot11\cdot12\right)+\left(11\cdot12\cdot13-11\cdot12\cdot13\right)+...+\left(498\cdot499\cdot500-498\cdot499\cdot500\right)+499\cdot500\cdot501-9\cdot10\cdot11\\ 3A=499\cdot500\cdot501-9\cdot10\cdot11\\ 3A=124999500-990\\ 3A=124998510\\ A=41666170\)
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