Lời giải:
Gọi tử số là $T$

\(T=(1-\frac{1}{6})+(1-\frac{2}{7})+(1-\frac{3}{8})+....+(1-\frac{88}{93})\)

\(=\frac{5}{6}+\frac{5}{7}+\frac{5}{8}+....+\frac{5}{93}=5(\frac{1}{6}+\frac{1}{7}+...+\frac{1}{93})\)

Gọi mẫu số là $M$
\(M=\frac{-1}{2}(\frac{1}{6}+\frac{1}{7}+....+\frac{1}{93})\)

Do đó:
\(C=\frac{5(\frac{1}{6}+\frac{1}{7}+...+\frac{1}{93})}{\frac{-1}{2}(\frac{1}{6}+\frac{1}{7}+...+\frac{1}{93})}=\frac{5}{\frac{-1}{2}}=-10\)

Ta có: \(A=1.3+2.4+3.5+4.6+...+99.101+100.102\)

\(A=1.\left(1+2\right)+2.\left(2+2\right)+3.\left(3+2\right)+4.\left(4+2\right)+....+99.\left(99+2\right)+100.\left(100+2\right)\)

\(A=\left(1^2+2^2+3^2+4^2+...+99^2+100^2\right)+\left(2+4+6+8+...+198+200\right)\)Đặt \(B=1^2+2^2+3^2+4^2+5^2+...+99^2+100^2\)

\(\Rightarrow B=\left(1^2+2^2+3^2+4^2+5^2+...+99^2+100^2\right)-2^2.\left(1^2+2^2+3^2+4^2+5^2+....+49^2+50^2\right)\)Tính dãy tổng quát \(C=1^2+2^2+3^2+4^2+5^2+...+n^2\)

\(C=1\left(0+1\right)+2\left(1+1\right)+3.\left(2+1\right)+4.\left(3+1\right)+5\left(4+1\right)+...+n\left[\left(n-1\right)+1\right]\)

\(C=\left[1.2+2.3+3.4+4.5+...+\left(n-1\right).n\right]+\left(1+2+3+4+5+....+n\right)\)

\(C=n.\left(n+1\right).\left[\left(n-1\right):3+1:2\right]=n.\left(n+1\right).\left(2n+1\right):6\)

Áp dụng vào B ta được:

\(B=100.101.201:6-4.50.51.101:6=166650\)

\(\Rightarrow A=166650+\left(200+2\right).100:2\)

\(\Rightarrow A=166650+10100=176750\)

Vậy A = 176750

Chúc bạn học tốt!!

c= - 1/10 nha

Giải:

\(\dfrac{\left(1-\dfrac{1}{6}\right)+\left(1-\dfrac{2}{7}\right)+\left(1-\dfrac{3}{8}\right)+...+\left(1-\dfrac{88}{93}\right)}{\dfrac{-1}{12}-\dfrac{1}{14}-\dfrac{1}{16}-...-\dfrac{1}{186}}\)   

Gọi dãy là A,phần tử là B. Ta có:

B=\(\left(1-\dfrac{1}{6}\right)+\left(1-\dfrac{2}{7}\right)+\left(1-\dfrac{3}{8}\right)+...+\left(1-\dfrac{88}{93}\right)\) 

B=\(\dfrac{5}{6}+\dfrac{5}{7}+\dfrac{5}{8}+...+\dfrac{5}{93}\) 

B=5.\(\left(\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+...+\dfrac{1}{93}\right)\) 

B=5.\(\left[\dfrac{2}{2}.\left(\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+...+\dfrac{1}{93}\right)\right]\) 

B=5.\(\left[2.\left(\dfrac{1}{12}+\dfrac{1}{14}+\dfrac{1}{16}+...+\dfrac{1}{186}\right)\right]\)

B=10.\(\left(\dfrac{1}{12}+\dfrac{1}{14}+\dfrac{1}{16}+...+\dfrac{1}{186}\right)\) 

⇒A=\(\dfrac{10.\left(\dfrac{1}{12}+\dfrac{1}{14}+\dfrac{1}{16}+...+\dfrac{1}{186}\right)}{\dfrac{-1}{12}+\dfrac{-1}{14}+\dfrac{-1}{16}+...+\dfrac{-1}{186}}\)

⇒A=-10

Chúc bạn học tốt!

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