35+335+3335+....+333...35(có 100 c/s 3)

=35+35+3x100+35+3x1100+....+3x111...00(99 c/s 1)+35

=35x100+3x(100+1100+...+111....100(99 c/s 1))

=3500+3x(1.100+11.10+....+111...1(99 c/s 1).10)

=3500+3 x ((1+11+..+111...1(99 c/s 1))x100)

Chịu thôi , đến đoạn này tớ bí !!!

Sorry nha!

trieu dang sao lại nói phúc như thế , mình cứ **** cho phúc thì sao 

A = \(\dfrac{1}{\sqrt{3}+\sqrt{5}}+\dfrac{1}{\sqrt{5}+\sqrt{7}}+\dfrac{1}{\sqrt{7}+\sqrt{9}}+...+\dfrac{1}{\sqrt{97}+\sqrt{99}}\)

= \(\dfrac{1}{2}\left(\sqrt{5}-\sqrt{3}+\sqrt{7}-\sqrt{5}+\sqrt{9}-\sqrt{7}+...+\sqrt{99}-\sqrt{97}\right)\)

= \(\dfrac{1}{2}\left(\sqrt{99}-\sqrt{3}\right)\)

B = 35 + 335 + 3335 + ... + 3333...(99 số 3)35

= 33 + 2 + 333 + 2 + 3333 + 2 + ... + 333...33 + 2

= 2 . 99 + (33 + 333 + 3333 + ... + 333...3)

= 198 + \(\dfrac{1}{3}\)(99 + 999 + 9999 + ... + 999...99)

= 198 + \(\dfrac{1}{3}\)(10² - 1 + 10³ - 1 + 10⁴ - 1 + ... + 10¹⁰⁰ - 1)

= \(\left(\dfrac{10^{101}-10^2}{27}\right)+165\)

đặt B=35 + 335 + 3335 + ..........(33.....335)
B-2.99=33+333+3333+...+(333....333)

(số cuối có 99 chữ số 3)

3.(B-2.99)=99+999+9999+....+(999...999...

(số cuối có 99 chữ số 9)

3(B-198)+99=100+1000+10000+....+(100..... (có 99 số 0)

3(B-198)+99=10^2+10^3+...+10^99
=(10^100-10^2)/9 (áp dụng công thức tính CSN)

=>3(B-198)+99=(10^100-10^2)/9

\(M=\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right).....\left(\frac{1}{100}-1\right)\left(\frac{1}{121}-1\right)=\frac{-3}{4}.\frac{-8}{9}.....\frac{-99}{100}.\frac{-120}{121}\)

\(M=\frac{-1.3}{2.2}.\frac{-2.4}{3.3}.....\frac{-9.11}{10.10}.\frac{-10.12}{11.11}=\frac{-1}{2}.\frac{-12}{11}=\frac{12}{22}=\frac{6}{11}\)

\(S=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{99}\)

\(S=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{9.11}\right)\)

\(S=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{9}-\frac{1}{11}\right)\)

\(S=\frac{1}{2}\left(1-\frac{1}{11}\right)\)

\(S=\frac{5}{11}\)

\(Q=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2013.2015}\)

\(Q=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2013.2015}\right)\)

\(Q=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2013}-\frac{1}{2015}\right)\)

\(Q=\frac{1}{2}\left(1-\frac{1}{2015}\right)\)

\(Q=\frac{1007}{2015}\)

~ Đấng Ed :) ~