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. 1155 nha bạn 

12 + 52 + 62 = 1 + 25 + 36 = 62

22 + 32 + 72= 4 + 9 + 49 = 62

Vậy 12 + 52 + 62 = 22 + 32 + 72

= 50

Nho t ick nha

     \(\left(102+82+62+42+22\right)-\left(12+32+53+72+92\right)\)

\(=102+82+62+42+22-12-32-52-72-92\)

\(=\left(102-92\right)+\left(82-72\right)+\left(62-52\right)+\left(42-32\right)+\left(22-12\right)\)

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

\(=10.5\)

\(=50\)

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

\(B=1+7+7^2+7^3+...+7^{150}\)

\(7B=7.\left(1+7+7^2+7^3+...+7^{150}\right)\)

\(7B=7+7^2+7^3+7^4+...+7^{151}\)

\(7B-B=\left(7+7^2+7^3+7^4+...+7^{151}\right)-\left(1+7+7^2+7^3+...+7^{150}\right)\)

\(6B=\left(7^{151}-1\right)\)

\(B=\left(7^{151}-1\right):6\)

  B      = 1 + 7 +  7² + ...+ 7¹⁵⁰

7.B     =      7  + 7²+.....+ 7¹⁵⁰ + 7¹⁵¹

7B - B =     7¹⁵¹  - 1

6B      =     7¹⁵¹ - 1

   B     = \(\dfrac{7^{151}-1}{6}\)