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

a:

Số số hạng trong dãy M là:

(1002-12):10+1=100(số)

=>Sẽ có 50 cặp (1002;992); (982;972);....;(22;12) có hiệu bằng 10

\(M=1002-992+982-972+...+22-12\)

\(=\left(1002-992\right)+\left(982-972\right)+...+\left(22-12\right)\)

\(=10+10+...+10\)

=10*50=500

b: \(N=\left(202+182+...+42+22\right)-\left(192+172+...+32+12\right)\)

\(=\left(202-192\right)+\left(182-172\right)+...+\left(22-12\right)\)

=10+10+...+10

=10*10=100

a) \(=\left(127+73\right)^2=200^2=40000\)

b) \(=18^8-\left(18^8-1\right)=1\)

c) \(=\left(100+99\right)\left(100-99\right)+\left(98+97\right)\left(98-97\right)+...+\left(2+1\right)\left(2-1\right)\)

\(=100+99+98+97+...+2+1=5050\)

d) biến đổi thành \(20^2-19^2+18^2-17^2+..+2^2-1^2\)

rồi giải ra như trên

\(S=1^2+3^2+5^2+\cdots+199^2\)

\(=\left(1^2+2^2+3^2+4^2+\cdots+200^2\right)-\left(2^2+4^2+\cdots+200^2\right)\)

\(=\left(1^2+2^2+\cdots+200^2\right)-2^2\left(1^2+2^2+\cdots+100^2\right)\)

\(=\frac{200\left(200+1\right)\left(2\cdot200+1\right)}{6}-4\cdot\frac{100\left(100+1\right)\left(2\cdot100+1\right)}{6}\)

\(=\frac{200\cdot201\cdot401}{6}-\frac23\cdot100\cdot101\cdot201\)

\(=100\cdot67\cdot401-100\cdot101\cdot134\)

\(=100\cdot67\left(401-101\cdot2\right)=6700\cdot199=1333300\)

\(1,=15+37+52-17-37=50\)

\(2,=21-32+12-32=-40\)

\(\text{1.(15+37)+(52-17-37)}\\ =15+37+52-17-37\\ =\left(15+52-17\right)+\left(37-37\right)\\ =50+0=50\)

\(\text{2.(21-32)-(-12+32)}\\ =21-32+12-32\\ =\left(21+12\right)+\left(-32-32\right)\\ =33+-64=-31\)

A. 1155 nha bạn