Ta có : \(S=\frac{989898.89-898989.98}{2^3+3^4+4^5+...+2014^{2015}}\)

\(=\frac{98\cdot10101\cdot89-89\cdot10101\cdot98}{2^3+3^4+4^5+...+2014^{2015}}\)

\(=\frac{10101\cdot\left(98\cdot89-89\cdot98\right)}{2^3+3^4+4^5+....+2014^{2015}}\)

\(=\frac{10101\cdot0}{2^3+3^4+4^5+....+2014^{2015}}=0\)

Vậy \(S=0\)

\(S=\frac{989898.89-898989.98}{2^3+3^4+4^5+...+2014^{2015}}\)

\(=\frac{98\cdot10101\cdot89-89\cdot10101\cdot98}{2^3+3^4+4^5+...+2014^{2015}}\)

\(=\frac{10101\cdot\left(98\cdot89-89\cdot98\right)}{2^3+3^4+4^5+...+2014^{2015}}\)

\(=\frac{10101\cdot0}{2^3+3^4+4^5+...+2014^{2015}}\)

\(=0\)

( - 2 )²⁰¹⁶ - ( - 2 )⁰

S = ( 1 - \(\dfrac{1}{2^2}\))(1-\(\dfrac{1}{3^2}\))(1-\(\dfrac{1}{4^2}\))....(1-\(\dfrac{1}{50^2}\))

S = \(\dfrac{2^2-1}{2^2}\).\(\dfrac{3^2-1}{3^2}\).\(\dfrac{4^2-1}{4^2}\)...\(\dfrac{50^2-1}{50^2}\)

Vì em lớp 6 nên phải làm thêm bước này nữa:

Ta có

n² - 1 = n² - n + n - 1 = (n² - n) + (n - 1) = n(n-1) + (n-1) =(n-1)(n+1)

Áp dụng công thức vừa chứng minh trên vào tổng S ta có:

S = \(\dfrac{\left(2-1\right)\left(2+1\right)}{2^2}\).\(\dfrac{\left(3-1\right)\left(3+1\right)}{3^2}\)....\(\dfrac{\left(50-1\right)\left(50+1\right)}{50^2}\)

S = \(\dfrac{1.3}{2^2}\).\(\dfrac{2.4}{3^2}\)......\(\dfrac{49.51}{50^2}\)

S = \(\dfrac{\left(3.4.5.6....49\right)^2.1.2.50.51}{\left(3.4.5.6...49\right)^2.2.2.50.50}\)

S = \(\dfrac{1}{2}\) . \(\dfrac{51}{50}\)

S = \(\dfrac{51}{100}\)

Em cảm ơn cô ạ1

kho..................lam............................tich,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,minh..........................troi........................ret............................wa.................ung ho minh.................hu....................hu..............hu................hat..............hat....................s

Vì S  chia hết cho 2 nhưng không chia hết cho 4 => S  không là SCP 

Vậy S không là số chính phương 

~HỌC TỐT NHÉ

Ta có: D = 1 + 4 + 4² + 4³ + .... + 4¹⁰⁰

=> 4D = 4 + 4² + 4³ + .... + 4¹⁰¹

=> 4D - D = 4¹⁰¹ - 1

=> 3D = 4¹⁰¹ - 1

=> D = 4¹⁰¹ - 1/3