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

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

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

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

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

uses crt;

var n,i:longint;

s:real;

{------------ham-tinh-giai-thua---------------------}

function gthua(x:longint):real;

var i:longint;

gt:real;

begin

gt:=1;

for i:=1 to x do

gt:=gt*i;

gthua:=gt;

end;

{------------chuong-trinh-chinh------------------}

begin

clrscr;

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

s:=0;

for i:=1 to n do 

  s:=s+gthua(i);

writeln(s:0:0);

readln;

end.

\(S=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right).....\left(1-\frac{1}{2016}\right)\)

\(=\left(\frac{2}{2}-\frac{1}{2}\right)\left(\frac{3}{3}-\frac{1}{3}\right)\left(\frac{4}{4}-\frac{1}{4}\right).....\left(\frac{2016}{2016}-\frac{1}{2016}\right)\)

\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{2015}{2016}=\frac{1}{2016}\)

\(S=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{2015}{2016}\)

\(S=\frac{1\cdot2\cdot3\cdot...\cdot2015}{2\cdot3\cdot4\cdot...\cdot2016}\)

\(S=\frac{1}{2016}\)

S=\(\left(1-\frac{1}{2}\right)x\left(1-\frac{1}{3}\right)x\left(1-\frac{1}{4}\right)x..x\left(1-\frac{1}{2016}\right)\)

S=\(\frac{1}{2}x\frac{2}{3}x\frac{3}{4}x...x\frac{2015}{2016}\)

S=1-\(\frac{2015}{2016}=\frac{1}{2016}\)

\(S=1\cdot2+2\cdot3+3\cdot4+...+99\cdot100\\ 3S=1\cdot2\cdot3+2\cdot3\cdot3+3\cdot3\cdot4+...+3\cdot99\cdot100\\ 3S=1\cdot2\cdot3+2\cdot3\cdot\left(4-1\right)+3\cdot4\cdot\left(5-2\right)+...+99\cdot100\cdot\left(101-98\right)\\ 3S=1\cdot2\cdot3+2\cdot3\cdot4-1\cdot2\cdot3+....+99\cdot100\cdot101-98\cdot99\cdot100\\ 3S=99\cdot100\cdot101\\ S=\dfrac{99\cdot100\cdot101}{3}=33\cdot100\cdot101=3300\cdot101=333300\)

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

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

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

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

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

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=s:)

S=(1+2)+(2^2+2^3)+(2^4+2^5)+....+(2^99+2^100)

S=3+3.2^2+3.2^4+.....+3.2^99

S=3.(2^2+2^4+.....+2^99)

Vì 3 chia hết 3=>3.(2^2+2^4+....+2^99)

=>S chia hết 3

2S=2+2^2+2^3+2^4+.....+2^101

2S-S=(2+2^2+2^3+2^4+....+2^101)-(1+2+2^2+2^3+2^4+....+2^100)

S=2^101-1

S+1=2^101-1+1=2^101

=>x=101

tích đúng cho mình nha

\(S=3+\frac{3}{2}+\frac{3}{2^2}+....+\frac{3}{2^9}\)

\(S\cdot\frac{1}{3}=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\)

\(S\cdot\frac{2}{3}=2+1+\frac{1}{2}+...+\frac{1}{2^8}\)

\(S\cdot\frac{2}{3}-S\cdot\frac{1}{3}=2+1+\frac{1}{2}+...+\frac{1}{2^8}-1-\frac{1}{2}-...-\frac{1}{2^9}\)

\(S\cdot\frac{1}{3}=2-\frac{1}{2^9}\)

\(S=\left(2-\frac{1}{2^9}\right):\frac{1}{3}\)

\(S=\left(2-\frac{1}{2^9}\right)\cdot3\)

\(S=6-\frac{3}{2^9}\)

\(S=\frac{6\cdot2^9-3}{2^9}\)