Answer
$$y' = \sum^{\infty}_{n=0} (n+1)^2x^{n}$$
- The coefficient of $x^n$ in the general term is: $(n+1)^2$
- The first four terms are:
$$y' = 1^2x^0 + 2^2x^1 + 3^2x^2 + 4^2x^3$$
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$$y'' = \sum^{\infty}_{n=0} (n+2)^2(n+1)x^{n}$$
- The coefficient of $x^n$ in the general term is: $(n+2)^2(n+1)$
- The first four terms are:
$$y'' = 2^2 \times 1x^0 + 3^2 \times 2x^1 + 4^2 \times 3 x^2 + 5^2 \times 4 x^3$$
Work Step by Step
1. Find the derivative of $nx^n$:
$$(nx^n)' = n^2x^{n-1}$$
- Therefore:
$$y' = \sum^{\infty}_{n=1} n^2x^{n-1}$$
- Write the expression so it involves $x^n$:
$$y' = \sum^{\infty}_{n=0} (n+1)^2x^{n}$$
- The coefficient of $x^n$ in the general term is: $(n+1)^2$
- The first four terms are:
$$y' = 1^2x^0 + 2^2x^1 + 3^2x^2 + 4^2x^3...$$
2. Find the derivative of $(n+1)^2x^n$
$$((n+1)^2x^n)' = (n+1)^2nx^{n-1}$$
- Therefore:
$$y'' = \sum^{\infty}_{n=1} (n+1)^2nx^{n-1}$$
- Write the expression so it involves $x^n$:
$$y'' = \sum^{\infty}_{n=0} (n+2)^2(n+1)x^{n}$$
- The coefficient of $x^n$ in the general term is: $(n+2)^2(n+1)$
- The first four terms are:
$$y'' = 2^2 \times 1x^0 + 3^2 \times 2x^1 + 4^2 \times 3 x^2 + 5^2 \times 4 x^3...$$
