A=1²/1.2 . 2²/2.3 . 3²/3.4 . 4²/4.5 . 5²/5.6

⇒1.1/1.2 . 2.2/2.3 . 3.3/3.4 . 4.4/4.5 . 5.5/5.6

⇒1.2.3.4.5/1.2.3.4.5 . 1.2.3.4.5/2.3.4.5.6

⇒1 . 1/6 =1/6.

Vậy A=1/6

B=2²/1.3 . 3²/2.4 . 4²/3.5 . 5²/4.6

⇒2.2/1.3 . 3.3/2.4 . 4.4/3.5 . 5.5/4.6

⇒2.2.3.3.4.4.5.5/1.3.2.4.3.5.4.6 =48.

Vậy B=48.

a,M=2^0-2^1+2^2-2^3+2^4-2^5+.....+2^2012

2M=2^1-2^2+2^3-2^4+2^5-2^5+......-2^2012+2^2013

3M=2^0+2^2013

M=(2^0+2^2013)÷3

Vậy.......

b,N=3-3^2+3^3-3^4+3^5-3^6+.....+3^2011-3^2012

3N=3^2-3^3+3^4-3^5+3^6-3^7+......+3^2012-3^2013

4N=3-3^2013

N=(3-3^2013)÷4

Vậy........

K tao nhé ko lên lớp tao đánh m😈😈😈

Bt dễ thế mà ko làm dc😂😂😂😂😂

Ta có: \(A=1\cdot3\cdot3+3\cdot5\cdot5+\cdots+97\cdot99\cdot99\)

\(=1\cdot3^2+3\cdot5^2+\cdots+97\cdot99^2\)

\(=3^2\left(3-2\right)+5^2\left(5-2\right)+\cdots+99^2\left(99-2\right)\)

\(=\left(3^3+5^3+\cdots+99^3\right)-2\left(3^2+5^2+\cdots+99^2\right)\)

\(=\left(1^3+3^3+5^3+\cdots+99^3\right)-2\left(1^2+3^2+5^2+\cdots+99^2\right)+1\)

Đặt \(B=1^3+3^3+\cdots+99^3\)

\(=\left(1^3+2^3+\cdots+100^3\right)-\left(2^3+4^3+\cdots+100^3\right)\)

\(=\left(1+2+\cdots+100\right)^2-2^3\left(1^3+2^3+\cdots+50^3\right)\)

\(=\left(100\cdot\frac{101}{2}\right)^2-8\cdot\left(1+2+\cdots+50\right)^2\)

\(=\left(50\cdot101\right)^2-8\cdot\left(50\cdot\frac{51}{2}\right)^2\)

\(=5050^2-8\cdot1275^2=12497500\)

Đặt \(C=1^2+3^2+5^2+\cdots+99^2\)

\(=\left(1^2+2^2+\cdots+100^2\right)-\left(2^2+4^2+\cdots+100^2\right)\)

\(=\frac{100\left(100+1\right)\left(2\cdot100+1\right)}{6}-2^2\left(1^2+2^2+\cdots+50^2\right)\)

\(=\frac{100\cdot101\cdot201}{6}-4\cdot\frac{50\left(50+1\right)\left(2\cdot50+1\right)}{6}\)

\(=50\cdot101\cdot67-\frac23\cdot50\cdot51\cdot101=338350-171700=166650\)

Ta có: A=B-2C+1

=12497500-2*166650+1

=12164201

Bài làm

   \(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right).....\left(1-\frac{1}{100}\right)\)

= \(\left(\frac{2}{2}-\frac{1}{2}\right).\left(\frac{3}{3}-\frac{1}{3}\right).\left(\frac{4}{4}-\frac{1}{4}\right).....\left(\frac{100}{100}-\frac{1}{100}\right)\)

= \(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{99}{100}\)

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

= \(\frac{1.1.1.....1}{1.1.1.....100}\)

= \(\frac{1}{100}\)

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cảm ơn bạn THẦN CHẾT nha!!!!!!!!

có bài nào khó thì bạn giải giúp mk nhé

\(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+...+\dfrac{2}{98.100}\\ =\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{98}-\dfrac{1}{100}\\ =\dfrac{1}{2}-\dfrac{1}{100}\\ =\dfrac{49}{100}\)

\(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+....+\dfrac{2}{98.100}\)\(=\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+....+\dfrac{1}{98}-\dfrac{1}{100}\)

                                                   \(=\dfrac{1}{2}-\dfrac{1}{100}=\dfrac{49}{100}\)

q=1/3; u1=2/3

\(S_{100}=\dfrac{\dfrac{2}{3}\cdot\left(\dfrac{1}{3^{100}}-1\right)}{\dfrac{1}{3}-1}=-\dfrac{1}{3^{100}}+1=\dfrac{-1+3^{100}}{3^{100}}\)