Bài 1 :

\(S=1.3+3.5+5.7+...+99.101=3+15+35+...9999\)

Ta thấy :

\(3=2^2-1\)

\(15=4^2-1\)

\(35=6^2-1\)

.....

\(9999=100^2-1\)

\(\Rightarrow S=2^2+4^2+...+100^2-\left(1\right).\left(\left(100-2\right):2+1\right)\)

\(\Rightarrow S=\dfrac{100.\left(100+1\right)\left(2.100+1\right)}{6}-51\)

\(\Rightarrow S=\dfrac{100.101.201}{6}-51=338299\)

nhanh len nhé mik đang cần gấp ai lam trước mik tích cho

Bài 6 :

\(C=1^2+2^2+...+100^2=\dfrac{100.\left(100+1\right)\left(2.100+1\right)}{6}=\dfrac{100.101.201}{6}=338350\)

Bài 9 :

\(S=1^2+2^2+3^2+...+99^2=\dfrac{99.\left(99+1\right)\left(2.99+1\right)}{6}=\dfrac{99.100.199}{6}=328350\)

A:

uses crt;

var t,i,n:integer;

begin

clrscr;

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

t:=0;

for i:=1 to n do

t:=t+i;

writeln(t);

readln;

end.

B:

uses crt;

var a:array[1..100]of integer;

i,n,t:integer;

begin

clrscr;

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

for i:=1 to m do

begin

write('A[',i,']='); readln(a[i]);

end;

t:=0;

for i:=1 to m do 

  t:=t+a[i];

writeln(t);

readln;

end.

\(aS=a+a^2+a^3+...+a^{n+1}\)

\(\Leftrightarrow S\left(a-1\right)=a^{n+1}-1\)

hay \(S=\dfrac{a^{n+1}-1}{a-1}\)

Program HOC24;

var i,n: integer;

S:longint;

begin

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

s:=0;

for i:=1 to n do s:=s+i*i;

write('S= ',s);

readln

end.