a) 1 + 2 + 3 + 4 + .... + n = 4950

=> n(n + 1)/2 = 4950

=> n(n + 1) = 9900

<=> n(n + 1) = 99 . 100

=> n = 99

b) 2 + 4 + 6 + .... + 2n = 210

=> 2(1 + 2 + 3 + ... + n) = 210

=> 1 + 2 + 3 + ... + n = 105

=> n(n + 1)/2 = 105

=> n(n + 1) = 210

=> n(n + 1) = 14.15

=> n = 14

c) 1 + 3 + 5 + ... + (2n - 1) = 225

=> [(2n - 1 - 1):2 + 1].(2n - 1 + 1) : 2 = 225

=> n.n = 225

=> n² = 15²

=> n = 15

\(\left(n+6\right)⋮\left(n+2\right)\)

\(\Rightarrow\left(n+2+4\right)⋮\left(n+2\right)\)

\(\Rightarrow4⋮n+2\)

\(\Rightarrow n+2\inƯ\left(4\right)=\left\{-4;-2;-1;1;2;4\right\}\)

\(\Rightarrow n\in\left\{-6;-4;-3;-1;0;2\right\}\)

\(\left(25n+3\right)⋮53\)

\(\Rightarrow\left(25n+3-53\right)⋮53\)

\(\Rightarrow\left(25n-50\right)⋮53\)

\(\Rightarrow25\left(n-2\right)⋮53\)

\(\text{Mà 25 không chia hết cho 53 nên }n-2⋮53\)

\(\Rightarrow n=53k+2\)

Sử dụng: 

\(A^3+B^3+C^3-3ABC=\left(A+B+C\right)\left(A^2+B^2+C^2-AB-BC-AC\right)\) (1)

Áp dụng vào bài:

\(\left(a-1\right)^3+\left(b-2\right)^3+\left(c-3\right)^3-3\left(a-1\right)\left(b-2\right)\left(c-3\right)\)

\(=\left(a-1+b-2+c-3\right)\)[ \(\left(a-1\right)^2+\left(b-2\right)^2+\left(c-3\right)^2\)

\(+\left(a-1\right)\left(b-2\right)+\left(a-1\right)\left(c-3\right)+\left(b-2\right)\left(c-3\right)\)]

<=> \(0-3\left(a-1\right)\left(b-2\right)\left(c-3\right)=0\)

( vì \(a-1+b-2+c-3=a+b+c-6=6-6=0\))

<=> \(\left(a-1\right)\left(b-2\right)\left(c-3\right)=0\)

<=>  a = 1 hoặc b = 2 hoặc c = 3.

Không mất tính tổng quát: g/s : a = 1

Khi đó: b + c =5

Ta có:  \(T=\left(b-2\right)^{2n+1}+\left(c-3\right)^{2n+1}\)

\(=\left(b-2+c-3\right).A\)

\(=\left(b+c-5\right).A\)

\(=0.A=0\)

Với \(A=\left(b-2\right)^{2n}-\left(b-2\right)^{2n-1}\left(c-3\right)+\left(b-2\right)^{2n-2}\left(c-3\right)^2-...+\left(c-3\right)^{2n}\)

Tương tự b = 2; c= 3 thì T = 0.

Vậy T = 0.

Chọn C