1. The problem asks to understand the difference between the two series expressions given: 2. The first series is the Taylor series expansion of $\sin(2x^3)$: $$\sin(2x^3) = \sum_{k=0}^\infty (-1)^k \frac{(2x^3)^{2k+1}}{(2k+1)!} = \sum_{k=0}^\infty (-1)^k \frac{2^{2k+1} x^{6k+3}}{(2k+1)!}$$ 3. The second series is the result of multiplying the entire function $\sin(2x^3)$ by $x$: $$x \sin(2x^3) = x \sum_{k=0}^\infty (-1)^k \frac{2^{2k+1} x^{6k+3}}{(2k+1)!} = \sum_{k=0}^\infty (-1)^k \frac{2^{2k+1} x^{6k+4}}{(2k+1)!}$$ 4. The key difference is the power of $x$ inside the summation: - In $\sin(2x^3)$, the power of $x$ is $6k+3$. - In $x \sin(2x^3)$, the power of $x$ is increased by 1 to $6k+4$ because of the multiplication by $x$. 5. This means multiplying by $x$ shifts the power of each term in the series by one, effectively changing the function's behavior and graph. 6. In summary, the difference is the multiplication by $x$ which increases the exponent of $x$ in every term of the series by 1. Final answer: $$x \sin(2x^3) = \sum_{k=0}^\infty (-1)^k \frac{2^{2k+1} x^{6k+4}}{(2k+1)!}$$ is the series for $\sin(2x^3)$ multiplied by $x$, shifting powers of $x$ by one.