\(vt=1+2015+2015^2+2015^3+2015^4+2015^5+2015^6+2015^7\)

\(=\left(1+2015\right)+\left(2015^2+2015^3\right)+\left(2015^4+2015^5\right)+\left(2015^6+2015^7\right)\)

\(=1\left(1+2015\right)+2015^2\left(1+2015\right)+2015^4\left(1+2015\right)+2015^6\left(1+2015\right)\)

\(=\left(2015+1\right)\left(1+2015^2+2015^4+2015^6\right)\)

\(=2016\left(1+2015^2+2015^4+2015^6\right)\)

\(=2016\left[\left(1+2015^2\right)+\left(2015^4+2015^6\right)\right]\)
\(=2016\left[1\left(1+2015^2\right)+2015^{2014}\left(1+2015^2\right)\right]=vp\left(đpcm\right)\)

\(=2016\left(1+2015^{2014}\right)\left(1+2015^{2012}\right)\)

cái chỗ =vp(đpcm ở dòng dưới nhé mk gõ nhầm)

Ta có : \(\dfrac{1}{2^2}\)<\(\dfrac{1}{1.2}\); \(\dfrac{1}{3^2}\)<\(\dfrac{1}{2.3}\);.....;\(\dfrac{1}{2016^2}\)<\(\dfrac{1}{2015.2016}\)

⇒ A = \(\dfrac{1}{2^2}\)+\(\dfrac{1}{3^2}\)+...+\(\dfrac{1}{2016^2}\)< \(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+...+\(\dfrac{1}{2015.2016}\)

⇒ A = \(\dfrac{1}{2^2}\)+\(\dfrac{1}{3^2}\)+...+\(\dfrac{1}{2016^2}\) < 1 - \(\dfrac{1}{2016}\)= \(\dfrac{2015}{2016}\) (ĐCPCM)

\(S=\left(1+3+3^2+3^3\right)+\left(3^4+3^5+3^6+3^7\right)+...+\left(3^{2012}+3^{2013}+3^{2014}+3^{2015}\right)\)

\(=\left(1+3+9+27\right)+3^4.\left(1+3+3^2+3^3\right)+...+3^{2012}.\left(1+3+3^2+3^3\right)\)

\(=40+3^4.40+...+3^{2012}.40\)

\(=40.\left(1+3^4+...+3^{2012}\right)\)

\(=10.4.\left(1+3^4+...+3^{2012}\right)\text{ chia hết cho 10}\)

=> S chia hết cho 10 (đpcm).

chtt

SCSH: (3²⁰¹⁵- 1) : 2 = 0

Tổng: (3²⁰¹⁵+ 1) : 2 = 2

Hk tốt,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,

k nhé

chưng minh mà anh

S = 1 + 3 + 3² + .......... + 3²⁰⁰⁸ + 3²⁰⁰⁹

= ( 1 + 3 ) + ( 3² + 3³ ) + ............. + ( 3²⁰⁰⁸ + 3²⁰⁰⁹ )

= 4 + 3²( 1 + 3 ) + ............ + 3²⁰⁰⁸( 1 + 3 )

= 4 + 4 . 3² + .......... + 4 . 3²⁰⁰⁸

= 4( 1 + 3² +......... + 3²⁰⁰⁸ ) chia hết cho 4

KL:......