1. We are given $f(x)=\sum_{k=1}^{10} kx^k.$ So, $f(x)=\sum_{k=1}^{10} kx^k.$

2. First compute $f(2)$: $f(2)=\sum_{k=1}^{10} k2^k.$

3. Differentiate $f(x)$ termwise: $f'(x)=\sum_{k=1}^{10} k\cdot kx^{k-1}=\sum_{k=1}^{10} k^2x^{k-1}.$ Hence, $f'(2)=\sum_{k=1}^{10} k^2 2^{k-1}.$

4. Now compute $2f(2)+f'(2).$ We have $2f(2)=2\sum_{k=1}^{10} k2^k=\sum_{k=1}^{10} k2^{k+1}.$ Therefore, $2f(2)+f'(2)=\sum_{k=1}^{10} k2^{k+1}+\sum_{k=1}^{10} k^2 2^{k-1}.$ Take $2^{k-1}$ common: $2f(2)+f'(2)=\sum_{k=1}^{10} 2^{k-1}(4k+k^2)=\sum_{k=1}^{10} 2^{k-1}k(k+4).$

5. Observe a smarter route using differentiation of $x^2f(x)$: Since $x^2f(x)=\sum_{k=1}^{10} kx^{k+2},$ its derivative is $\frac{d}{dx}[x^2f(x)]=2xf(x)+x^2f'(x).$ At $x=2$, $\frac{d}{dx}[x^2f(x)]_{x=2}=4f(2)+4f'(2).$ This is not directly the desired form, so instead use the identity termwise: $2\cdot k2^k+k^22^{k-1}=2^{k-1}(k^2+4k)=2^{k-1}k(k+4).$ Now compute the sum explicitly near the highest power.

6. Let us check whether the expression can be written as $119\cdot 2^n+1$. We compute the sum directly: $2f(2)+f'(2)=\sum_{k=1}^{10} 2^{k-1}k(k+4).$ Termwise values are:

• $k=1$: $2^0\cdot1\cdot5=5$
• $k=2$: $2^1\cdot2\cdot6=24$
• $k=3$: $2^2\cdot3\cdot7=84$
• $k=4$: $2^3\cdot4\cdot8=256$
• $k=5$: $2^4\cdot5\cdot9=720$
• $k=6$: $2^5\cdot6\cdot10=1920$
• $k=7$: $2^6\cdot7\cdot11=4928$
• $k=8$: $2^7\cdot8\cdot12=12288$
• $k=9$: $2^8\cdot9\cdot13=29952$
• $k=10$: $2^9\cdot10\cdot14=71680$

Adding, $5+24+84+256+720+1920+4928+12288+29952+71680=121857.$ So, $2f(2)+f'(2)=121857.$

7. Now compare with $119\cdot 2^n+1.$ Then $119\cdot 2^n=121856.$ So, $2^n=\frac{121856}{119}=1024=2^{10}.$ Hence, $n=10.$

8. Comparison with stored answer: Stored correct answer is $10$, which matches our result.