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

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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

b) Tại x=14 thì:\(B\left(x\right)=x^5-15x^4+16x^3-29x^2+13x\)

\(=x^5-\left(x+1\right)x^4+\left(x+2\right)x^3-\left(2x+1\right)x^2+x\left(x-1\right)\)

\(=x^5-x^5-x^4+x^4+2x^3-2x^3-x^2+x^2-x=-x=-14\)

a) A(x)=1

cậu giúp mình nốt phần kia đc k cậu

\(C=12+22+32+...+992+1002\)

\(=1+2\left(1+1\right)+3\left(2+1\right)+...+99\left(98+1\right)+100\left(99+1\right)\)

\(=1+1.2+2+2.3+3+...+98.99+99+99.100+100\)

\(=\left(1.2+2.3+3.4+...+99.100\right)+\left(1+2+3+...+99+100\right)\)\(=333300+5050=338350\)