Let $$f(x) = \sum_{k=1}^{10} k \cdot x^k$$, $$x \in \mathbb{R}$$, if $$2f(2) + f'(2) = 119 \cdot 2^n + 1$$ then $$n$$ is equal to _____.

$$f(x) = \sum_{k=1}^{10} kx^k$$, so $$f'(x) = \sum_{k=1}^{10} k^2 x^{k-1}$$.

Computing numerically:

$$f(2) = 2 + 8 + 24 + 64 + 160 + 384 + 896 + 2048 + 4608 + 10240 = 18434$$

$$f'(2) = 1 + 8 + 36 + 128 + 400 + 1152 + 3136 + 8192 + 20736 + 51200 = 84989$$

$$2f(2) + f'(2) = 36868 + 84989 = 121857$$

$$119 \times 2^n + 1 = 121857$$

$$119 \times 2^n = 121856$$

$$2^n = 1024 = 2^{10}$$

Therefore $$n = \mathbf{10}$$.