\(A=2^1+2^2+2^3+...+2^{2016}\)

\(\Rightarrow A=2\left(1+2^1+2^2\right)+2^4\left(1+2^1+2^2\right)...+2^{2014}\left(1+2^1+2^2\right)\)

\(\Rightarrow A=2.7+2^4.7...+2^{2014}.7\)

\(\Rightarrow A=7\left(2+2^4...+2^{2014}\right)⋮7\)

\(\Rightarrow dpcm\)

a, 2.(x – 5)+7 = 77

<=> 2.(x – 5) = 70 <=> x – 5 = 35 <=> x = 40

b,  x - 1 3 - 3 5 : 3 4 + 2 . 2 3 = 14

<=> x - 1 3 - 3 + 2 4 = 14

<=>  x - 1 3 = 14 + 3 - 16 = 1

<=> x – 1 = 1 <=> x = 2

c,  1 + 2 + 2 2 + 2 3 + . . . + 2 2016 = 2 x - 1 - 1

Đặt: A = 1 + 2 + 2 2 + 2 3 + . . . + 2 2016 => 2A =  2 + 2 2 + 2 3 + . . . + 2 2017

=> 2A – A = ( 2 + 2 2 + 2 3 + . . . + 2 2017 ) – ( 1 + 2 + 2 2 + 2 3 + . . . + 2 2016 )

=> A =  2 2017 - 1

Ta có:  1 + 2 + 2 2 + 2 3 + . . . + 2 2016 = 2 x - 1 - 1 =>  2 2017 - 1 =  2 x - 1 - 1 => x = 2018

d,  5 2 x - 3 - 2 . 5 2 = 5 2 . 3

<=>  5 2 x - 3 = 5 2 . 3 + 5 2 . 2

<=>  5 2 x - 3 = 5 2 . ( 3 + 2 )

<=>  5 2 x - 3 = 5 3

<=> 2x – 3 = 3 => x = 3

a) \(A=1+2+2^2+...+2^{80}\)

\(2A=2+2^2+2^3+...+2^{81}\)

\(2A-A=2+2^2+2^3+...+2^{81}-1-2-2^2-...-2^{80}\)

\(A=2^{81}-1\)

Nên A + 1 là:

\(A+1=2^{81}-1+1=2^{81}\)

b) \(B=1+3+3^2+...+3^{99}\)

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

\(3B-B=3+3^2+3^3+...+3^{100}-1-3-3^2-...-3^{99}\)

\(2B=3^{100}-1\)

Nên 2B + 1 là:

\(2B+1=3^{100}-1+1=3^{100}\)

2) 

a) \(2^x\cdot\left(1+2+2^2+...+2^{2015}\right)+1=2^{2016}\)

Gọi:

\(A=1+2+2^2+...+2^{2015}\)

\(2A=2+2^2+2^3+...+2^{2016}\)

\(A=2^{2016}-1\)

Ta có:

\(2^x\cdot\left(2^{2016}-1\right)+1=2^{2016}\)

\(\Rightarrow2^x\cdot\left(2^{2016}-1\right)=2^{2016}-1\)

\(\Rightarrow2^x=\dfrac{2^{2016}-1}{2^{2016}-1}=1\)

\(\Rightarrow2^x=2^0\)

\(\Rightarrow x=0\)

b) \(8^x-1=1+2+2^2+...+2^{2015}\)

Gọi: \(B=1+2+2^2+...+2^{2015}\)

\(2B=2+2^2+2^3+...+2^{2016}\)

\(B=2^{2016}-1\)

Ta có:

\(8^x-1=2^{2016}-1\)

\(\Rightarrow\left(2^3\right)^x-1=2^{2016}-1\)

\(\Rightarrow2^{3x}-1=2^{2016}-1\)

\(\Rightarrow2^{3x}=2^{2016}\)

\(\Rightarrow3x=2016\)

\(\Rightarrow x=\dfrac{2016}{3}\)

\(\Rightarrow x=672\)

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