C= (1 - \(\frac{1}{2^2}\))+(1 - \(\frac{1}{3^2}\) )+(1 - \(\frac{1}{4^2}\))+.......+(1 - \(\frac{1}{100^2}\))

  =98 - (\(\frac{1}{2^2}\)+\(\frac{1}{3^2}\)+\(\frac{1}{4^2}\)+........+\(\frac{1}{100^2}\))

=> C< 98            bn xem lai nha hinh nhu de sai  phai cong den \(\frac{9999}{10000}\)

Uk hinh nhu sai

A=-1-2...-98

A=-98-97...-2-1

=>2A=[(-98-1)+(-97-2)...(-51-49)]-50

=>2A=(-99-99...-99)-50(có 98 số hạng trừ 50)

Vậy A=(-99.98/2)-50=4901

Sửa đề: 100-98+96-94+...+12-10+8-6+4

Số số hạng trong dãy số 6;8;10;...;98;100 là:

\(\frac{100-6}{2}+1=\frac{94}{2}+1=47+1=48\) (số)

Ta có: 100-98+96-94+...+12-10+8-6+4

=(100-98)+(96-94)+...+(12-10)+(8-6)+4

=2+2+...+2+4

\(=2\cdot\frac{48}{2}+4=48+4=52\)

Thánh Gióng

a: Ta có: \(A=2+2^2+2^3+\cdots+2^{2025}\)

=>\(2A=2^2+2^3+2^4+\cdots+2^{2026}\)

=>\(2A-A=2^2+2^3+2^4+\cdots+2^{2026}-2-2^2-\cdots-2^{2025}\)

=>\(A=2^{2026}-2\)

b:Sửa đề: \(B=1+5+5^2+\cdots+5^{150}\)

=>\(5B=5+5^2+5^3+\cdots+5^{151}\)

=>\(5B-B=5+5^2+5^3+\cdots+5^{151}-1-5-5^2-\cdots-5^{150}\)

=>\(4B=5^{151}-1\)

=>\(B=\frac{5^{151}-1}{4}\)

c: Ta có: \(C=3+3^2+3^3+\ldots+3^{1000}\)

=>\(3C=3^2+3^3+3^4+\cdots+3^{1001}\)

=>\(3C-C=3^2+3^3+\cdots+3^{1001}-3-3^2-\cdots-3^{1000}\)

=>\(2C=3^{1001}-3\)

=>\(C=\frac{3^{1001}-3}{2}\)

\(\frac{1}{95}\frac{1}{95}\)

là sao ???

\(\frac{x-1}{99}+\frac{x-2}{98}+\frac{x-5}{95}=3+\frac{1}{99}+\frac{1}{98}+\frac{1}{95}\)

\(\frac{x-1}{99}+\frac{x-2}{98}+\frac{x-5}{95}=\frac{2765070}{921690}+\frac{9310}{921690}+\frac{9405}{921690}+\frac{9702}{921690}\)

\(\frac{x-1}{99}+\frac{x-2}{98}+\frac{x-5}{95}=\frac{2793487}{921690}\)

\(BCNN\left(99,98,95\right)=921690\Rightarrow x=101\)

= \(\frac{125875}{125871}\)= 1.000031779

hUÝUYG