\(a,A=1\cdot2+2\cdot3+...+98\cdot99\\ 3A=1\cdot2\cdot3+2\cdot3\cdot3+3\cdot4\cdot3+...+98\cdot99\cdot3\\ 3A=1\cdot2\cdot3+2\cdot3\cdot\left(4-1\right)+3\cdot4\left(5-2\right)+...+98\cdot99\left(100-97\right)\\ 3A=1\cdot2\cdot3-1\cdot2\cdot3+2\cdot3\cdot4-2\cdot3\cdot4+3\cdot4\cdot5-...-97\cdot98\cdot99+98\cdot99\cdot100\\ 3A=98\cdot99\cdot100=970200\\ A=323400\)

\(b,B=1^2+2^2+3^3+...+98^2\\ B=1\left(2-1\right)+2\left(3-1\right)+3\left(4-1\right)+...+98\left(99-1\right)\\ B=\left(1\cdot2+2\cdot3+3\cdot4+...+98\cdot99\right)-\left(1+2+...+98\right)\\ B=323400-\left[\left(98+1\right)\left(98-1+1\right):2\right]\\ B=323400-4851=318549\\ c,C=1\cdot99+2\left(99-1\right)+3\left(99-2\right)+...+98\left(99-97\right)+99\left(99-98\right)\\ C=1\cdot99+2\cdot99-1\cdot2+3\cdot99-2\cdot3+...+98\cdot99-97\cdot98+99\cdot99-98\cdot99\\ C=99\left(1+2+...+99\right)-\left(1\cdot2+2\cdot3+...+98\cdot99\right)\\ C=99\left[\left(99+1\right)\left(99-1+1\right):2\right]-323400\\ C=490050-323400=166650\)

https://hoc24.vn/cau-hoi/a-tinh-tong-a1223349899b-su-dung-ket-qua-cau-a-tinh-b122232972982c-su-dung-ket-qua-cau-a-tinh-c1992983979829.2030286199021

:vv hỏi hoài z?

\(A=1\cdot2+2\cdot3+...+151\cdot152\)

\(=1\left(1+1\right)+2\left(1+2\right)+...+151\left(1+151\right)\)

\(=\left(1+2+3+...+151\right)+\left(1^2+2^2+...+151^2\right)\)

\(=\dfrac{151\left(151+1\right)}{2}+\dfrac{151\left(151+1\right)\left(2\cdot151+1\right)}{6}\)

\(=151\cdot76+\dfrac{151\cdot152\cdot303}{6}\)

\(=151\cdot76+151\cdot7676=1170552\)

\(C=2\cdot4+4\cdot6+...+2024\cdot2026\)

\(=2\cdot2\left(1\cdot2+2\cdot3+...+1012\cdot1013\right)\)

\(=4\left[1\left(1+1\right)+2\left(1+2\right)+...+1012\left(1+1012\right)\right]\)

\(=4\left[\left(1+2+...+1012\right)+\left(1^2+2^2+...+1012^2\right)\right]\)

\(=4\left[1012\cdot\dfrac{1013}{2}+\dfrac{1012\left(1012+1\right)\left(2\cdot1012+1\right)}{6}\right]\)

\(=4\left[506\cdot1013+345990150\right]\)

\(=1386010912\)

\(M=1^2+2^2+...+2024^2\)

\(=\dfrac{2024\left(2024+1\right)\cdot\left(2\cdot2024+1\right)}{6}\)

\(=2024\cdot2025\cdot\dfrac{4049}{6}\)

=2765871900

\(N=1^3+2^3+...+100^3\)

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

\(=\left[\dfrac{100\left(100+1\right)}{2}\right]^2\)

\(=\left[50\cdot101\right]^2=5050^2\)

\(Q=1^3+2^3+...+2024^3\)

\(=\left(1+2+3+...+2024\right)^2\)

\(=\left[\dfrac{2024\left(2024+1\right)}{2}\right]^2\)

\(=\left[1012\left(2024+1\right)\right]^2\)

\(=2049300^2\)

\(1\cdot2+2\cdot3+3\cdot4+...+n\left(n+1\right)\\ =\dfrac{1}{3}\left[1\cdot2\cdot3+2\cdot3\cdot3+...+3n\left(n+1\right)\right]\\ =\dfrac{1}{3}\left[1\cdot2\left(3-0\right)+2\cdot3\left(4-1\right)+...+n\left(n+1\right)\left(n+2-n+1\right)\right]\\ =\dfrac{1}{3}\left[1\cdot2\cdot3-1\cdot2\cdot3+2\cdot3\cdot4-...-\left(n-1\right)n\left(n+1\right)+n\left(n+1\right)\left(n+2\right)\right]\\ =\dfrac{n\left(n+1\right)\left(n+2\right)}{3}\)

2:

a: =>101y+5050=5555

=>101y=505

=>y=5

b: =>3^y+1=3^5

=>y+1=5

=>y=4