Đặt P=3^1-1+3^2-1+3^3-1+3^4-1+...+3^n-1

=>P=3⁰+3¹+3²+3³+...+3^n-1

=>3.P=3¹+3²+3³+3⁴+...+3ⁿ

=>3.P-P=3¹+3²+3³+3⁴+...+3ⁿ-3⁰-3¹-3²-3³-...-3^n-1

=>2.P=3ⁿ-3⁰

=>2.P=3ⁿ-1

=>\(P=\frac{3^n-1}{2}\)

Lại có: S=1+2+5+14+...+\(\frac{3^{n-1}+1}{2}\)

=>\(S=\frac{3^{1-1}+1}{2}+\frac{3^{2-1}+1}{2}+\frac{3^{3-1}+1}{2}+\frac{3^{4-1}+1}{2}+...+\frac{3^{n-1}+1}{2}\)

=>\(S=\frac{3^{1-1}+1+3^{2-1}+1+3^{3-1}+1+3^{4-1}+1+...+3^{n-1}+1}{2}\)

=>\(S=\frac{\left(3^{1-1}+3^{2-1}+3^{3-1}+3^{4-1}+...+3^{n-1}\right)+\left(1+1+1+1+...+1\right)}{2}\)

=>\(S=\frac{P+1.n}{2}\)

=>\(S=\frac{\frac{3^n-1}{2}+n}{2}\)

=>\(S=\frac{\frac{3^n-1}{2}+\frac{2n}{2}}{2}\)

=>\(S=\frac{\frac{3^n-1+2n}{2}}{2}\)

=>\(S=\frac{3^n-1+2n}{4}\)

nhìn cái cuối là biết quy luật đó bạn :))

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

\(S=\frac{\left(3^0+3^1+....+3^{n-1}\right)+\left(1+1+1+...+1\right)}{2}\left(\text{ có n c/s 1}\right)\)

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

chỗ 3⁰+3¹+...+3^n-1 bn tự tính :))

S = 1 + 2 + 3 + ... + n

S = n(n + 1) : 2

2S = n(n + 1)

2S ⋮ 2 

=> n(n + 1) ⋮ 2

#include <bits/stdc++.h>

using namespace std;

double s,a;

int i,n;

int main()

{

cin>>a;

s=0;

n=0;

while (s<=a) 

{

n=n+1;

s=s+1/(n*1.0);

}

cout<<n;

return 0;

}

Bài 1: 

uses crt;

var n,i:integer;

s:real;

begin

clrscr;

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

s:=0;

for i:=1 to n do 

  s:=s+1/(2*i+1);

writeln(s:4:2);

readln;

end.

bài 1

BÀI 3

uses crt;
var a: array[1..100] of integer;
i,n,max,s,j: integer;
begin
clrscr;
writeln(' nhap so phan tu cua day'); readln(n);
for i:=1 to n do
begin
writeln('a[',i,']'); readln(a[i]);
end;
max:=a[1];
s:=0;
for i:=1 to n do
begin
if max<a[i] then max:=a[i];
s:=s+a[i];
end;
writeln('so lon nhat trong day tren la:',max);
writeln('tong bang:',s);
writeln('so nguyen to trong mang la:');
j:=1;
for i:=1 to n do
if a[i]>1 then
begin
repeat
inc(j);
until (a[i] mod j=0);
if j>(a[i] div 2) then writeln(a[i]);
j:=1;
end;
readln
end.

1:

uses crt;

var n,i,t:integer;

begin

clrscr;

readln(n);

t:=0;

for i:=1 to n do

t:=t+i*i;

write(t);

readln;

end.

2

program bt2;

var i,n,t:integer;

begin

readln(n);

s:=0;

for i:=1 to n do

if i mod 2 = 1 then s:=s+i;

readln;

end.