Calculating matrix $\ [T+I]^{10}$

Question

Let $\ T: \mathbf R^n \rightarrow \mathbf R^n $, $\ T(x_1,...,x_n) = (0,(n-1)x_1,(n-2)x_2,...,x_{n-1}) $

Now suppose $\ n =3 $ and $\ I$ is the identity transformation of $\ \mathbf R^3 $. I need to calculate $\ [T+I]_E^{10} $

I was instructed to use the binom so $\ (T+I)^{10} = T^{10} + 10T^9I + 45 T^8 I^2 + 120 T^7 I^3 + 210 T^6I^4 + 252T^5I^5 + 210T^4I^6 + 120 T^3 I^7 + 45 T^2 I ^8 + 10 T I^9 + I^{10} = \\T^{10} + 10(T + T^9) + 45(T^2 + T^8) + 120(T^7 + T^3) + 210(T^6+T^4) + 252T^5 + I^{10} $

Which is still to long to calculate... How can I make it simpler? $\ T+I $ can't be diagonalized.

Answer 1

Hint:

Calculate $T^3$.

(I know it's a really short answer, but I don't want to spoil everything, it should become clear after that)