Lời giải:

Ta có:
\(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2015^2}\)

\(S> \frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2015.2016}\)

\(\Leftrightarrow S> \frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{2016-2015}{2015.2016}\)

\(\Leftrightarrow S> \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2015}-\frac{1}{2016}\)

\(\Leftrightarrow S> \frac{1}{2}-\frac{1}{2016}=\frac{1007}{2016}\)

--------------------------

\(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{2015^2}\)

\(S< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{2014}{2015}\)

\(\Leftrightarrow S< \frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{2015-2014}{2014.2015}\)

\(\Leftrightarrow S< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-....+\frac{1}{2014}-\frac{1}{2015}\)

\(\Leftrightarrow S< 1-\frac{1}{2015}=\frac{2014}{2015}\)

Vậy ta có đpcm.

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)

Ta có:\(B=1+2015+2015^2+...+2015^{99}\)

=>\(2015B=2015+2015^2+2015^3+...+2015^{100}\)

=>\(2015B-B=2014B=2015^{100}-1\)

=>\(2014B+1=2015^{100}=\left(2015^{50}\right)^2\)

Vì 2014B + 1 là bình phương của một số tự nhiên

Vậy 2014B + 1 là số chính phương

Ta có : \(B=1+2015+2015^2+...+2015^{99}\)

\(\Rightarrow2015B=2015+2015^2+2015^3+...+2015^{100}\)

\(\Rightarrow2015B-B=2014B=2015^{100}-1\)

\(\Rightarrow2014B+1=2015^{100}=\left(2015^{50}\right)^2\)

Vì : \(2014B+1\) là bình phương của một số tự nhiên

Vậy \(2014B+1\) là số chính phương

khó wá

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

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

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

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

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

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

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

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