tổng quát: với mọi số n \(\ne\) 0;ta luôn có:

\(\frac{2}{\left(n-1\right)n\left(n+1\right)}=\frac{1}{n\left(n-1\right)}-\frac{1}{n\left(n+1\right)}\)

Đặt \(S=\frac{1}{1.2.3}+\frac{1}{2.3.4}+....+\frac{1}{98.99.100}\)

\(\Rightarrow2S=\frac{2}{1.2.3}+\frac{2}{2.3.4}+......+\frac{2}{98.99.100}\)

\(\Rightarrow2S=\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\left(\frac{1}{98.99}-\frac{1}{99.100}\right)\)

\(\Rightarrow2S=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}=\frac{1}{1.2}-\frac{1}{99.100}=\frac{1}{2}-\frac{1}{9900}=\frac{4949}{9900}\)

\(\Rightarrow S=\frac{4949}{9900}:2=\frac{4949}{19800}\)

Vậy S=4949/19800

Đề sai rồi nha bạn

Phải là:

\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...........+\frac{1}{98.99.100}\) chứ

Ai đồng ý với mình thì ***** nha

Ta có: 16a=25b=30c mà a,b,c nhỏ nhất nên 16a=25b=30c=BCNN(16;25;30) hay 16a=25b=30c=1200 =>a=1200/16=75 =>b=1200/25=48 =>c=1200/30=40

Câu a)
\(A=2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\)
\(=\left(2^{100}+2^{99}+2^{98}+2^{97}+...+2^2+2\right)-2\left(2^{99}+2^{97}+2^{95}+...+2^3+2\right)\)
\(=\left(2^{100}+2^{99}+2^{98}+2^{97}+...+2^2+2\right)-\left(2^{100}+2^{98}+2^{96}+...+2^4+2^2\right)\)
\(=2^{99}+2^{97}+2^{95}+...+2^3+2\)
\(=\frac{2^2\cdot\left(2^{99}+2^{97}+2^{95}+...+2^3+2\right)-\left(2^{99}+2^{97}+2^{95}+...+2^3+2\right)}{3}\)
\(=\frac{\left(2^{101}+2^{99}+2^{97}+...+2^5+2^3\right)-\left(2^{99}+2^{97}+2^{95}+...+2^3+2\right)}{3}\)
\(=\frac{2^{101}-2}{3}\)

\(2B=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{2015.2016.2017}\)

\(2B=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{2.4}+...+\frac{1}{2015.2016}-\frac{1}{2016.2017}\)

\(2B=\frac{1}{1.2}-\frac{1}{2016.2017}\)

\(B=\frac{\frac{1}{1.2}-\frac{1}{2016.1017}}{2}\)

nói chung là: đặt tổng đó là E

suy ra:

2E=1/1.2-1/99.100

=> E=(1/1.2-1/99.100):2=(1/1.2-1/99.100).1/2

vậy k=2

2 đúng rùi

còn cách giải dài lắm

mk lười ghi

4F = 1.2.3(4-0) + 2.3.4(5-1) +3.4.5.(6-2) +......+ k.(k+1)(k+2)[(k+3) - (k-1)]

    = 1.2.3.4 - 0 + 2.3.4.5 - 1.2.3.4 + 3.4.5.6 - 2.3.4.5 + ......+ k(k+1)(k+2)(k+3)- (k-1)k(k+1)(k+2) = k(k+1)(k+2)(k+3)

F = \(\frac{k\left(k+1\right)\left(k+2\right)\left(k+3\right)}{4}\)

=> 4F + 1 =\(k\left(k+1\right)\left(k+2\right)\left(k+3\right)+1\)=[k(k+3)][(k+1)(k+2)] +1 =(k²+3k)(k²+3k+2) + 1

               = (k²+3k)² +2(k²+3k) +1 = (k²+3k+1)²

=> 4F + 1 là số chình phương

Ta có : \(k\left(k+1\right)\left(k+2\right)=\frac{1}{4}k\left(k+1\right)\left(k+2\right).4\)

\(=\frac{1}{4}k\left(k+1\right)\left(k+2\right)\left[\left(k+3\right)-\left(k-1\right)\right]\)

\(=\frac{1}{4}k\left(k+1\right)\left(k+2\right)\left(k+3\right)-\frac{1}{4}k\left(k+1\right)\left(k+2\right)\left(k-1\right)\)

=> 4S = 1.2.3.4-0.1.2.3+2.3.4.5-1.2.3.4+...+k(k+1)(k+2)(k+3)-k(k+1)(k+2)(k-1)

\(=k\left(k+1\right)\left(k+2\right)\left(k+3\right)\)

=> \(4S+1=k\left(k+1\right)\left(k+2\right)\left(k+3\right)+1\)

\(=\left[k\left(k+3\right)\right]\left[\left(k+1\right)\left(k+2\right)\right]+1\)
\(=\left[\left(k^2+3k\right)\left(k^2+k+2k+2\right)\right]+1\)

Đặt \(t=k^2+3k\)

\(=>4S+1=t\left(t+2\right)+1\)

= \(t^2+2t+1\)

\(=\left(t+1\right)^2\)

\(=>4S+1=\left(k^2=3k\right)^2=>4S+1\) là số chính phương