S = 1 + 3 + 3² + 3³ +... + 3²⁰¹⁴

3S = 3 + 3² + 3³ + 3⁴ + ... + 3²⁰¹⁵

3S - S = ( 3 + 3² + 3³ + 3⁴ + ... + 3²⁰¹⁵) - (1 + 3 + 3² + 3³ +... + 3²⁰¹⁴)

2S = 3²⁰¹⁵ - 1

S = \(\dfrac{3^{2015}-1}{2}\)

S = (1/31+1/32+1/33+...+1/40) + (1/41 + 1/42 + ...+ 1/50) + (1/51 + 1/52+...+1/59+1/60)

Mà : (1/31+1/32+1/33+...+1/40) > 1/40 x 10 = 1/4 (gồm 10 số hạng)

Tương tự : (1/41 + 1/42 + ...+ 1/50) > 1/5 ; (1/51 + 1/52+...+1/59+1/60) > 1/6

S > 1/4 + 1/5 + 1/6.

Trong khi đó (1/4 + 1/5 + 1/6) > 3/5

=>S > 3/5 (1)

S = (1/31+1/32+1/33+...+1/40) + (1/41 + 1/42 + ...+ 1/50) + (1/51 + 1/52+...+1/59+1/60)

Mà : (1/31+1/32+1/33+...+1/40) < 1/31 x 10 = 10/30 = 1/3 (gồm 10 số hạng)

=> S < 4/5 (2)

Từ (1) và (2) => 3/5 <S<4/5

so sanh 2 vế nha

vế 1 chứng minh S>3/5

ta có:S=1/31+1/32+.......+1/60>10.1/40+10.1/50+10.1/60=1/4+1/5+1/6=37/60>3/5

vậy S>3/5

vế 2 chứng minh S<4/5

ta có:S=1/31+1/32+.....+1/60<10.1/30+10.1/40+10.1/50=1/3+1/4+1/5=47/60<4/5

vậy S<4/5

giúp mk với, mk tk

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

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

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

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

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

xin lỗi bài trên của mình làm sai

Ta có: 3A = 3.(1+3+3²+3³+...+3⁹⁹+3¹⁰⁰⁾ 

3A = 3+3²+3³+...+3¹⁰⁰+3¹⁰¹

Suy ra: 3A – A = (3+3²+3³+...+3¹⁰⁰+3¹⁰¹)−(1+3+3²+3³+...+3⁹⁹+3¹⁰⁰)

2A = 3¹⁰¹−1

⇒ A = 3¹⁰¹−1

             2               

Vậy A = 3¹⁰¹−1

                 2           

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