\(S=1+3+3^2+3^3+...+3^{2014}\)

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

\(3S-S=\left(3+3^2+3^3+3^4+...+2^{2015}\right)-\left(1+3+3^2+3^3+...+3^{2014}\right)\)

\(2S=3^{2015}-1\)

\(S=\frac{3^{2015}-1}{2}\)

toi chịu >:) nhắn cho vui thoi

\(S=1^2+2^2+3^2+...+n^2\)

\(=1.2-1+2.3-2+3.4-3+...+n\left(n+1\right)-n\)

\(=\left[1.2+2.3+3.4+...+n\left(n+1\right)\right]-\left(1+2+3+...+n\right)\)

Theo dạng tổng quát: \(1.2+2.3+3.4+...+n\left(n+1\right)=\frac{n\left(n+1\right)\left(n+2\right)}{3}\)

\(\Rightarrow S=\frac{n\left(n+1\right)\left(n+2\right)}{3}-\frac{n\left(n+1\right)}{2}\)

\(=\frac{2n\left(n+1\right)\left(n+2\right)}{6}-\frac{3n\left(n+1\right)}{6}\)

\(=\frac{2n\left(n+1\right)\left(n+2\right)-3n\left(n+1\right)}{6}\)

\(=\frac{n\left(n+1\right).\left[2\left(n+2\right)-3\right]}{6}=\frac{n\left(n+1\right)\left(2n+1\right)}{6}\)

Vậy \(S=\frac{n\left(n+1\right)\left(2n+1\right)}{6}\)

Ta có : \(S=1^2+2^2+3^2+...+\)\(n^2\)

\(\Rightarrow S=\frac{n.\left(n+1\right)\left(n+2\right)}{2}\)

Xin lỗi mình nhớ nhầm công thức : \(S=\frac{n\left(n+1\right)\left(2n+1\right)}{6}\)

S=2+4+6+...+98+100

S=\(\frac{\left[\left(\frac{100-2}{2}+1\right).\left(100+2\right)\right]}{2}=2550\)

S=1+2+3+4+...+2016+2017

S=\(\frac{\left(2017-1+1\right).\left(2017+1\right)}{2}=2035153\)

1.Số lượng số của S= (2017-1)+1=2017 số

tổng=(2016+1).(2016:2)+2017=2 035 153

2.Số lượng số của S=(100-2):2+1=50 số

tổng=(100+2).(50:2)=2 550

uses crt;

var i,n:longint;

s:real;

begin

clrscr;

write('Nhap n='); readln(n);

s:=0;

for i:=1 to n do 

  s:=s+sqr(i);

writeln(s:0:0);

readln;

end.

\(S=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2019}}\)

\(\Rightarrow2S=1+\frac{1}{2}+...+\frac{1}{2^{2018}}\)

\(\Rightarrow2S-S=\left(1+\frac{1}{2}+...+\frac{1}{2^{2018}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2019}}\right)\)

\(\Rightarrow S=1-\frac{1}{2^{2019}}\)

Ta có: \(S=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2019}}\)

\(\frac{1}{2}S=\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2020}}\)

\(\Rightarrow\frac{1}{2}S=S-\frac{1}{2}S=\frac{1}{2}-\frac{1}{2^{2020}}\Rightarrow S=1-\frac{1}{2^{2019}}.\)

\(\frac{1}{2}S=\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2020}}\)

\(S-\frac{1}{2}S=\frac{1}{2}-\frac{1}{2^{2020}}\)

\(\Rightarrow\frac{1}{2}S=\frac{1}{2}-\frac{1}{2^{2020}}\)

\(\Rightarrow S=\frac{1}{4}-\frac{1}{2^{2021}}\)

:) 

Ta có: S=2²+4²+6²+...+20²

            =(2.1)²+(2.2)²+(2.3)²+...+(2.10)²

            =2².1²+2².2²+2².3²+...+2².10²

            =2².(1+2²+3²+...+10²)

Mà 1²+2²+3²+...+10²=385 nên:

S=2².385

  =4.385

  =1540

Vậy S=1540

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