TK bài làm nhé:

Ta có: \(1^2+3^2+5^2+\cdots+99^2\)

\(=1^2+2^2+\cdots+100^2-\left(2^2+4^2+\cdots+100^2\right)\)

\(=\left(1^2+2^2+\cdots+100^2\right)-2^2\left(1^2+2^2+\cdots+50^2\right)\)

\(=\frac{100\left(100+1\right)\left(2\cdot100+1\right)}{6}-4\cdot\frac{50\cdot\left(50+1\right)\left(2\cdot50+1\right)}{6}\)

\(=\frac{100\cdot101\cdot201}{6}-\frac{2\cdot50\cdot51\cdot101}{3}=50\cdot101\cdot67-100\cdot17\cdot101\)

\(=101\cdot50\left(67-2\cdot17\right)=5050\cdot\left(67-34\right)=33\cdot5050\)

=166650

uses crt;

var i,n,s:longint;

begin

clrscr;

readln(n);

s:=0;

for i:=1 to n do 

  s:=s+sqr((2*i-1));

writeln(s);

readln;

end.

A. 1155 nha bạn 

\(A=1^2+2^2+3^2+....+10^2\\ A=1^{ }+\left(1+1\right)\cdot2+3\cdot\left(2+1\right)+.....+10\cdot\left(9+1\right)\\ A=1+2\cdot1+2+3\cdot2+3+....+10\cdot9+10\\ A=\left(1+2+3...+10\right)+\left(1\cdot2+3\cdot2+.....+10\cdot9\right)\)

Gọi 1+2+3+...+10 là P

Số số hạng là: (10 - 1) : 1 +1 = 10 (số)

P = (10+1) . 10 : 2 = 55 

P = 55

Gọi \(1\cdot2+2\cdot3+....+9\cdot10\)  là C

\(C=1\cdot2+2\cdot3+....+9\cdot10\\ 3\cdot C=1\cdot2\cdot3+2\cdot3\cdot3+....+9\cdot10\cdot3\\ 3\cdot C=1\cdot2\cdot3+2\cdot3\cdot\left(4-1\right)+....+9\cdot10\cdot\left(11-8\right)\\ 3\cdot C=1\cdot2\cdot3+2\cdot3\cdot4-1\cdot2\cdot3+.....+9\cdot10\cdot11-8\cdot9\cdot10\\ 3\cdot C=9\cdot10\cdot11\\ 3\cdot C=990\\ C=330\)

\(=>A=P+C\\ =>A=55+330\\ A=385\)

b)

\(B=5^2+10^2+15^2+...+50^2\\ B=5^2+\left(2\cdot5\right)^2+\left(3\cdot5\right)^2+....+\left(5\cdot10\right)^2\\ B=5^2+2^2\cdot5^2+3^2\cdot5^2+...+5^2\cdot10^2\\ B=5^2\cdot\left(1+2^2+3^2+....+10^2\right)\\ B=25\cdot\left(1+2^2+3^2+....+10^2\right)\)

\(\left(1+2^2+3^2+....+10^2\right)=A\)

\(=>B=25\cdot A\\ B=25\cdot385\\ B=9625\)

1

1

Sửa đề: \(P=2\cdot101+3\cdot100+4\cdot99+\cdots+99\cdot4+100\cdot3+101\cdot2\)

Ta có: \(P=2\cdot101+3\cdot100+4\cdot99+\cdots+99\cdot4+100\cdot3+101\cdot2\)

\(=2\left(2\cdot101+3\cdot100+4\cdot99+\cdots+51\cdot52\right)\)

\(=2\left\lbrack2\cdot\left(103-2\right)+3\left(103-3\right)+\cdots+51\left(103-51\right)\right\rbrack\)

\(=2\cdot\left\lbrack103\left(2+3+\cdots+51\right)-\left(2^2+3^2+\cdots+51^2\right)\right\rbrack\)

\(=2\cdot\left\lbrack103\cdot\left(51-2+1\right)\cdot\frac{\left(51+2\right)}{2}-\left(1^2+2^2+\cdots+51^2\right)+1^2\right\rbrack\)

\(=2\cdot\left\lbrack103\cdot50\cdot\frac{53}{2}-\frac{51\cdot\left(51+1\right)\left(2\cdot51+1\right)}{6}+1\right\rbrack\)

\(=2\cdot\left\lbrack103\cdot25\cdot53-\frac{51\cdot52\cdot103}{6}+1\right\rbrack=2\cdot\left\lbrack103\cdot25\cdot53-17\cdot26\cdot103+1\right\rbrack\)

=181900

Ta có: \(Q=2^2+3^2+\cdots+101^2\)

\(=1^2+2^2+3^2+\cdots+101^2-1\)

\(=101\left(101+1\right)\cdot\frac{\left(2\cdot101+1\right)}{6}-1=101\cdot102\cdot\frac{203}{6}-1\)

\(=101\cdot17\cdot203-1=348551-1=348550\)

P+Q

=181900+348550

=530450

bạn so sánh từng số hạng là được

hello các bạn mình đang có tâm trạng rất buồn😔😔😔😔😔